You are Qwen, created by Alibaba Cloud. You are a helpful assistant.

Reasoning language: Ukrainian
Reasoning effort: medium

<|im_start|>system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.

Reasoning language: Ukrainian
Reasoning effort: low<|im_end|>
<|im_start|>user
Обчисліть площу бічної поверхні правильної трикутної піраміди, сторона основи якої дорівнює 8 см, а апофема на 2 см більша за сторону основи піраміди.
<|im_end|>
<|im_start|>assistant
<think>
Задача: Обчислити площу бічної поверхні правильної трикутної піраміди, де сторона основи дорівнює 8 см, а апофема на 2 см більша за сторону основи.

Кроки:

1. **Знаходження апофеми (l):**
   - Апофема = сторона основи + 2 см = 8 см + 2 см = 10 см.

2. **Обчислення периметра основи (P):**
   - Основа — правильний трикутник, тому периметр P = 3 × сторона = 3 × 8 см = 24 см.

3. **Формула площі бічної поверхні (S):**
   - S = (P × l) / 2 = (24 см × 10 см) / 2 = 120 см².

Відповідь: Площа бічної поверхні піраміди становить 120 см².

Аналіз завершено. Переходжу до розв'язку.
</think>

Площа бічної поверхні правильної трикутної піраміди дорівнює 120 см².

**Пояснення:**
1. **Апофема (l):** 8 см (сторона основи) + 2 см = 10 см.
2. **Периметр основи (P):** 3 × 8 см = 24 см.
3. **Формула площі бічної поверхні:** $ S = \frac{P \cdot l}{2} = \frac{24 \cdot 10}{2} = 120 \, \text{см}^2 $.

**Відповідь:** 120 см².<|im_end|>

<|im_start|>system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.

Reasoning language: Ukrainian
Reasoning effort: medium<|im_end|>
<|im_start|>user
Обчисліть площу бічної поверхні правильної трикутної піраміди, сторона основи якої дорівнює 8 см, а апофема на 2 см більша за сторону основи піраміди.
<|im_end|>
<|im_start|>assistant
<think>
Гаразд, мені потрібно обчислити площу бічної поверхні правильної трикутної піраміди. Дано, що сторона основи дорівнює 8 см, а апофема на 2 см більша за сторону основи. Давайте розберемося крок за кроком.

Спочатку, правильною трикутною пірамідою називається піраміда, основа якої — правильний трикутник, а бічні ребра рівні. Площа бічної поверхні такої піраміди — це сума площ усіх бічних граней. Оскільки основа — правильний трикутник, всі три бічні грані є рівними рівнобедреними трикутниками. Тому площу бічної поверхні можна обчислити, помноживши площу однієї бічної грані на 3.

Формула для площі бічної поверхні правильної піраміди: S = (периметр основи × апофема) / 2. Апофема — це висота бічної грані, яка є висотою рівнобедреного трикутника (бічної грані). У цій задачі апофема на 2 см більша за сторону основи. Сторона основи дорівнює 8 см, тому апофема становить 8 + 2 = 10 см.

Периметр основи — це периметр правильного трикутника зі стороною 8 см. Периметр — це сума всіх сторін, тому 3 × 8 = 24 см.

Підставляючи значення до формули: S = (24 × 10) / 2 = 240 / 2 = 120 см². Отже, площа бічної поверхні піраміди дорівнює 120 квадратних сантиметрів.

Перевіряю, чи все логічно. Апофема — це висота бічної грані, а не висота піраміди. Тут використовується стандартна формула для площі бічної поверхні піраміди, яка полягає у множенні периметра основи на апофему та діленні на два. Оскільки всі три бічні грані однакові, це правильно. Периметр основи — 24 см, апофема — 10 см. 24 × 10 = 240, поділене на 2 дає 120. Всі кроки видають однаковий результат, тому відповідь має бути 120 см².

Але, можливо, я щось пропустив. Давайте ще раз переконаємось. Формула правильна? Так, для площі бічної поверхні піраміди вона є правильна. Оскільки кожна бічна грань — це трикутник з основою 8 см і висотою (апофемою) 10 см. Площа однієї грані — (8 × 10)/2 = 40 см². Три грані дають 3 × 40 = 120 см². Так, це підтверджує результат. Отже, відповідь правильна.

Інші можливі помилки? Наприклад, якщо було б запитано про висоту піраміди, а не апофему, але тут у задачі чітко сказано, що апофема на 2 см більша за сторону основи. Тому 8 + 2 = 10 см. Вірно. Тому що апофема — це висота бічної грані, а не висота піраміди. Тому формула застосовується правильно.

Отже, відповідь — 120 см².

Аналіз завершено. Переходжу до розв'язку.
</think>

Щоб обчислити площу бічної поверхні правильної трикутної піраміди, використаємо формулу:
$$
S = \frac{P \cdot a}{2},
$$
де $ P $ — периметр основи, $ a $ — апофема.

1. **Периметр основи** ($ P $):
   - Основа — правильний трикутник зі стороною $ 8 \, \text{см} $.
   - $ P = 3 \cdot 8 = 24 \, \text{см} $.

2. **Апофема** ($ a $):
   - За умовою, апофема на $ 2 \, \text{см} $ більша за сторону основи: $ a = 8 + 2 = 10 \, \text{см} $.

3. **Площа бічної поверхні**:
   $$
   S = \frac{24 \cdot 10}{2} = 120 \, \text{см}^2.
   $$

**Відповідь:** $ \boxed{120} $.<|im_end|>

<|im_start|>system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.

Reasoning language: English
Reasoning effort: medium<|im_end|>
<|im_start|>user
Обчисліть площу бічної поверхні правильної трикутної піраміди, сторона основи якої дорівнює 8 см, а апофема на 2 см більша за сторону основи піраміди.
<|im_end|>
<|im_start|>assistant
<think>
Okay, let's see. I need to find the lateral surface area of a regular triangular pyramid. The base edge is 8 cm, and the apothem is 2 cm larger than the base edge. Hmm, first, I should recall what a regular triangular pyramid is. It's a pyramid with a base that's an equilateral triangle, and all the lateral faces are congruent isosceles triangles. The apothem here probably refers to the slant height of the pyramid, right? Because in pyramids, the apothem is the distance from the apex to the midpoint of one of the base edges, which is the height of each triangular face. So the lateral surface area would be the sum of the areas of these three triangular faces.

Wait, the problem says the apothem is 2 cm larger than the side of the base. The base side is 8 cm, so the apothem is 8 + 2 = 10 cm? That seems a bit big, but maybe it's correct. Let me confirm. The apothem (slant height) is 10 cm. Each lateral face is a triangle with base 8 cm and height 10 cm. The area of one triangular face would be (base * height)/2 = (8 * 10)/2 = 40 cm². Since there are three such faces, the total lateral surface area would be 3 * 40 = 120 cm². Is that all?

Wait, hold on. Maybe I need to check if the apothem is indeed the slant height. In some contexts, apothem can refer to the distance from the center of the base to the midpoint of a side, which in a regular polygon is the same as the radius of the inscribed circle. But in pyramids, the apothem (or slant height) is the distance from the apex to the midpoint of a base edge. So in this problem, they specifically mention the apothem is 2 cm larger than the base edge. So if the base edge is 8 cm, the apothem is 10 cm. So the slant height is 10 cm. Therefore, each triangular face has area (8*10)/2 = 40, three faces give 120. That seems straightforward.

But let me make sure there's no confusion between apothem and slant height. In regular pyramids, the slant height is the apothem of the pyramid. So yes, the apothem here is the slant height. So the calculation should be correct. Alternatively, maybe the problem is using "apothem" differently? For example, in a regular pyramid, the apothem is the distance from the apex to the midpoint of a base edge, which is the slant height. So in that case, yes, the slant height is 10 cm. Therefore, the lateral surface area is 3*(1/2)*8*10 = 120 cm². So the answer should be 120 cm².

Wait, but let me check once more. The base is a regular triangle with side 8 cm. The apothem (slant height) is 10 cm. The formula for lateral surface area of a regular pyramid is (perimeter of base * slant height)/2. The perimeter of the base is 3*8=24 cm. So lateral surface area is (24*10)/2=120 cm². Yes, that's the same result. So that's correct. Therefore, the answer is 120 cm². I think that's it.
</think>

Для обчислення площі бічної поверхні правильної трикутної піраміди, де сторона основи дорівнює $ a = 8 \, \text{см} $, а апофема (скільки вона є висотою бічної грані) на $ 2 \, \text{см} $ більша за сторону основи ($ h = a + 2 = 10 \, \text{см} $):

1. **Площа однієї бічної грані**:
   $$
   S_{\text{грані}} = \frac{1}{2} \cdot a \cdot h = \frac{1}{2} \cdot 8 \cdot 10 = 40 \, \text{см}^2.
   $$

2. **Загальна площа бічної поверхні** (три грані):
   $$
   S_{\text{бічн}} = 3 \cdot S_{\text{грані}} = 3 \cdot 40 = 120 \, \text{см}^2.
   $$

**Відповідь:** $ \boxed{120} \, \text{см}^2 $.<|im_end|>

<|im_start|>system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.

Reasoning language: English
Reasoning effort: high<|im_end|>
<|im_start|>user
You’re given a competitive programming problem.

Analyze the maximum input constraints and identify the optimal algorithmic approach and data structures needed to process the largest possible test cases within the time and memory limits, then explain why your chosen implementation strategy is the most efficient solution. Please reason step by step about your solution approach, then provide a complete implementation in Python 3 that is thoroughly optimized for both speed and memory usage.

Your solution must read input from standard input (input()), write output to standard output (print()).
Do not include any debug prints or additional output.

Put your final solution within a single code block:
'''python
<your code here>
'''

# Problem

You're given an integer $$$n$$$. For every integer $$$i$$$ from $$$2$$$ to $$$n$$$, assign a positive integer $$$a_i$$$ such that the following conditions hold:

- For any pair of integers $$$(i,j)$$$, if $$$i$$$ and $$$j$$$ are coprime, $$$a_i \neq a_j$$$.
- The maximal value of all $$$a_i$$$ should be minimized (that is, as small as possible).

A pair of integers is called coprime if their greatest common divisor is $$$1$$$.

## Constraints
Time limit per test: 1.0 seconds
Memory limit per test: 256.0 megabytes

## Input Format
The only line contains the integer $$$n$$$ ($$$2 \le n \le 10^5$$$).

## Output Format
Print $$$n-1$$$ integers, $$$a_2$$$, $$$a_3$$$, $$$\ldots$$$, $$$a_n$$$ ($$$1 \leq a_i \leq n$$$).

If there are multiple solutions, print any of them.

## Examples
'''input
4
'''
'''output
1 2 1
'''
'''input
3
'''
'''output
2 1
'''
<|im_end|>
<|im_start|>assistant
<think>
Okay, let's tackle this problem. So the task is to assign a positive integer a_i to each i from 2 to n such that if i and j are coprime, their a_i's are different. And we need to minimize the maximum a_i. Hmm.

First, I need to understand the conditions. For any coprime pair (i,j), a_i must not equal a_j. So the main challenge is to assign colors (since a_i can be thought of as colors) such that coprime numbers have different colors. And the goal is to use as few colors as possible (since the maximum a_i is minimized).

Wait, but how do we model this? It's similar to a graph coloring problem where each number is a node, and there's an edge between two nodes if they are coprime. Then we need to find the minimum number of colors such that adjacent nodes (coprimes) have different colors. But this graph is huge for n up to 1e5, so we can't model it directly. So we need a smarter approach.

Let me think about the problem differently. The key is to assign colors in such a way that any two coprime numbers get different colors. So the colors are determined by some property of the numbers that ensures this.

What's the minimal number of colors needed? Let's think of small examples.

In the first example, n=4. The output is 1 2 1. Let's see:

i=2: a=1
i=3: a=2
i=4: a=1

Check coprimes:

2 and 3 are coprime (gcd 1), so a_2 (1) != a_3 (2) ✔️.
2 and 4: gcd 2, not coprime. So their colors can be same. So a_2 and a_4 are both 1. That's allowed.
3 and 4: coprime. a_3 is 2, a_4 is 1. Different. ✔️.

So the maximum a_i here is 2. How is this achieved?

Another example: n=3. Output is 2 1. So a_2=2, a_3=1. Since 2 and 3 are coprimes, their colors are different. The maximum is 2.

Wait, but why not assign a_2=1 and a_3=2? Then the maximum is 2 as well. So both options are possible. The problem allows any valid solution as long as the maximum is minimized.

So the minimal maximum is 2 for n=3, 2 for n=4. But wait, what about when n is larger?

Let's think about n=5. Let's see what's possible. Let's see the coprimes:

For i=2, coprimes are 3,5 (since 2 and 4 are not coprime, 2 and 5 are coprime).

So a_2 must differ from a_3 and a_5.

Similarly, a_3 must differ from a_2, a_4 (gcd(3,4)=1?), yes. Wait, 3 and 4 are coprime. So a_3 must differ from a_4.

Wait, 3 and 4: gcd is 1. So yes. So a_3 and a_4 must be different.

So how can we assign colors here?

Maybe the approach is to assign colors based on the smallest prime factor (spf) of each number. Because if two numbers are coprime, their spf's are different. Wait, no. For example, 6 and 5 are coprime. The spf of 6 is 2, spf of 5 is 5. So their spf's are different. But 6 and 3 are not coprime. So maybe the color can be the spf. Let's see.

If two numbers are coprime, their spf's must be different. Because if they share a prime factor, then they are not coprime. So if two numbers are coprime, their spf's are different. Therefore, assigning the color as the spf would ensure that coprime numbers have different colors. Because their spf's are different.

But wait, what about numbers that are primes themselves. For example, 5 and 7. Both are primes. So their spf's are 5 and 7. So colors would be 5 and 7. But that's a problem because the maximum color would be up to n, which is not optimal. For example, in the first example, n=4, the spf of 3 is 3, but the color assigned is 2. So this approach might not be the best.

Hmm, perhaps the minimal maximum color is determined by the maximum number of primes that are coprime. Wait, but primes are all coprime with each other. So if we have a set of primes, each must have a unique color. So the minimal number of colors needed is at least the number of primes in the set. But that's not possible for large n. For example, when n is 1e5, there are about 10^4 primes. So the maximum color would be 1e4, which is way higher than the examples given. So this approach can't be correct.

Wait, but the examples show that for n=4, the maximum color is 2. So perhaps there's a different approach. Let's think again.

Another idea: the color can be determined by the parity (even or odd) of the number. But that's not sufficient. For example, 2 and 3 are coprime. 2 is even, 3 is odd. So their colors would be different. 3 and 4 are coprime (3 is odd, 4 even). So colors differ. 2 and 4 are not coprime, same color. So maybe this works? Let's test for n=4.

If a_i is 1 if i is even, 2 if i is odd. Then:

a_2=1 (even), a_3=2 (odd), a_4=1 (even). So coprimes 2-3 (diff colors), 3-4 (diff colors). 2-4 (same color, allowed). So this works. The maximum is 2. Which matches the example. For n=3, a_2=1 (even), a_3=2 (odd). Which also matches the example. So maybe this approach works.

Wait, but what about numbers that are not coprime but have different parity. For example, 6 (even) and 9 (odd). They are not coprime (gcd 3). So their colors can be same. So that's allowed. But for coprime pairs, their parity must differ. So this approach ensures that all coprime pairs have different colors. Because if two numbers are coprime, one must be even and the other odd. Because if both are even, they share 2 as a factor. So they can't be coprime. So any coprime pair must consist of one even and one odd. Therefore, their colors (parity) are different. So this approach would work.

But wait, what about 3 and 5 (both odd). They are coprime, but their colors are the same (since they are both odd). So that's a problem. Because according to the parity approach, they would have the same color, which violates the condition. So the parity approach is invalid.

Oh right! Because 3 and 5 are both odd primes, they are coprime. So their colors must be different. So the parity approach fails here.

So that approach is incorrect. So how to handle this?

Hmm. So the parity approach works for even-odd pairs, but not for odd-odd coprimes.

So perhaps the color needs to be based on something else. Let's think again.

Another approach: group numbers by their smallest prime factor. For example, numbers with spf=2 (even numbers) get color 1. Numbers with spf=3 get color 2. Numbers with spf=5 get color 3, etc. But then, two coprime numbers would have different spf's, so their colors would be different. For example, 3 (spf 3) and 5 (spf 5) would have colors 2 and 3. So different. But 3 and 7 (spf 7) would have different colors. So this approach works. But the problem is that the maximum color would be the number of distinct primes up to n, which for n=1e5 is about 10^4, which is way larger than the examples. So this approach would not be optimal.

Wait, but the examples show that for n=4, the maximum color is 2. So perhaps there's a smarter way to group numbers into colors such that coprime numbers are in different groups, but the maximum color is minimized.

Another idea: the colors are determined by the parity of the number of prime factors. Wait, not sure.

Alternatively, perhaps the color is determined by the residue modulo some k. For example, if we can find a k such that any two coprime numbers have different residues mod k. But how to find such k?

Alternatively, think about the problem in terms of graph coloring. The graph is the coprime graph, and we need to find the chromatic number. But chromatic number is the minimum number of colors needed. However, for this graph, what is the chromatic number?

Wait, the coprime graph is a complement of the non-coprime graph. So two numbers are adjacent if they are coprime. So the problem is to color this graph with the minimal number of colors.

But what's the chromatic number of the coprime graph? For example, the chromatic number for the coprime graph on integers up to n.

Hmm. Let's think for small n.

For n=2: only 2. No edges. So color 1. But since the output is for 2 to n, which is just 2. So a_2 can be 1.

For n=3: 2 and 3 are coprime. So the graph is an edge between 2 and 3. So chromatic number is 2. So a_2 and a_3 must be different. Which matches the example.

For n=4: 2-3, 2-5 (but 5 is not in this case). Wait, for n=4, the numbers are 2,3,4. The coprime pairs are (2,3), (3,4). So the graph is 2 connected to 3, and 3 connected to 4. So the graph is a path of 3 nodes. The chromatic number is 2. So colors can be 1,2,1. Which matches the example.

For n=5: the coprimes include 2-3, 2-5, 3-4, 3-5, 4-5, 2-5, etc. So the graph is more connected. For example, 2 is connected to 3,5. 3 is connected to 2,4,5. 4 is connected to 3,5. 5 is connected to 2,3,4. So the graph is a complete graph on 2,3,5 (since 2,3,5 are pairwise coprime). So the chromatic number is 3. So the maximum color would be 3. So for n=5, the output would need to have a_2, a_3, a_5 all different. So how?

But how to model this? It seems that the chromatic number is the maximum number of primes in the set. Wait, but 2,3,5 are primes and pairwise coprime. So they form a clique of size 3. So the chromatic number is at least 3. So for n >=5, the chromatic number is at least 3. But how to color the rest of the numbers.

Alternatively, perhaps the chromatic number is 2 if the maximum number of primes in the set is 2, but higher otherwise.

But this seems complicated. Let's think differently.

Wait, the problem requires that for any two coprime numbers, their colors are different. So the color of a number must be different from the colors of all other numbers that are coprime with it.

So for a number x, all its coprimes must have different colors. But how to assign colors in a way that this is satisfied with minimal maximum color.

An alternative approach: assign colors based on the number's parity, but for primes, assign a color that's different from others. Wait, not sure.

Alternatively, perhaps the color can be determined by the parity of the number's smallest prime factor. For example, for even numbers (spf 2), color 1. For numbers with spf 3, color 2. For spf 5, color 3, etc. But again, this leads to high maximum colors.

But the examples show that for n=4, the maximum color is 2. So perhaps there's a pattern here. Let's look at the examples again.

In the first example, n=4. The output is 1 2 1. So a_2=1 (even), a_3=2 (prime), a_4=1 (even). So for even numbers, color 1. For primes (3), color 2. But 3 is a prime, and 3 is coprime with 2 and 4. So their colors are different.

In the second example, n=3. Output is 2 1. So a_2=2 (even), a_3=1 (prime). So primes are assigned color 1, even numbers color 2. Wait, but this contradicts the first example. So perhaps there's a different pattern.

Wait, perhaps the color is determined by the parity of the number of prime factors (including multiplicity) modulo some number. Not sure.

Alternatively, perhaps the color is determined by the residue of the number modulo the maximum number of primes in the set. But again, not sure.

Another idea: The problem requires that any two coprime numbers have different colors. So the color of a number x must be different from the colors of all numbers y where gcd(x,y)=1. So for x, all y that are coprime with x must have different colors.

But how to model this. Let's think for each x, the set of colors that x can take must not be present in any of its coprimes. So the color of x must be unique among all coprimes of x.

Wait, but that's not possible for all x. For example, if x is even, then any even number is not coprime with x. So x can share the same color with other even numbers. But any odd number coprime with x must have a different color.

So perhaps the color can be assigned based on whether the number is even or odd. But as discussed earlier, this fails for primes that are odd and coprime. Like 3 and 5.

So perhaps the color needs to be based on the parity and something else. For example, even numbers get color 1. For odd numbers, assign colors based on their parity in another way.

Wait, maybe for odd numbers, we can assign colors based on their residue modulo some k. For example, for even numbers, color 1. For odd numbers, color 2 if the number is a prime, and color 1 otherwise. But then, primes (like 3 and 5) would have color 2, which is different from each other. But 3 and 5 are coprimes, so their colors must be different. So this approach would work for primes. But what about composite odd numbers that are coprime with primes.

For example, 9 (composite, odd) and 5 (prime). They are coprime. So 9 would have color 1 (since it's composite), 5 color 2. So different. So that's okay. But what about 9 and 7 (prime). 9 is color 1, 7 is color 2. Different. So that's okay.

But what about two composite odd numbers that are coprime. For example, 9 and 25. They are coprime. 9 is color 1 (composite), 25 is color 1 (composite). So same color, which is invalid. So this approach would fail here.

So that's a problem. So this approach is not sufficient.

Hmm. So perhaps the problem requires a more sophisticated approach.

Another idea: The color of a number can be determined by the parity of the number, but for primes, the color is determined by another factor. For example, even numbers get color 1. For primes, assign colors in a way that primes are all different. But this would require a lot of colors, which is not optimal.

Alternatively, think of the problem in terms of grouping numbers into colors such that no two coprimes are in the same color. So for even numbers, color 1. For primes, color 2. For composite odd numbers, color 1. But then, primes (color 2) and even numbers (color 1) are different. So primes are coprime with even numbers. So their colors are different. But primes are coprime with each other. So all primes would have the same color (2), which is invalid. So this approach doesn't work.

Wait, but the primes are all coprime with each other. So all primes must have different colors. So the number of colors needed is at least the number of primes. But that's impossible for large n. So this suggests that the minimal maximum color is the number of primes, but that's not possible given the examples. So perhaps there's a different way to assign colors that doesn't require all primes to have unique colors.

Wait, but in the first example, n=4. The primes are 2 and 3. But a_2=1 and a_3=2. So they have different colors. So for primes, they are assigned different colors. But for n=5, primes are 2,3,5. So they would need three different colors. But how to assign colors to other numbers.

Alternatively, perhaps the color is determined by the smallest prime factor. For example:

- For even numbers (spf=2), color 1.
- For numbers with spf=3, color 2.
- For spf=5, color 3, etc.

In this case, any two coprime numbers must have different spf's. Because if two numbers are coprime, their spf's must be different. So their colors would be different. So this approach ensures that all coprimes have different colors. But the maximum color would be the number of primes up to n. For n=1e5, this is about 10^4, which is way larger than the examples. So this approach is not optimal.

But the examples show that for n=4, the maximum color is 2, which is the number of primes (2 and 3). So this approach would work for the examples. But for larger n, the maximum color would be higher.

But the problem requires the maximum a_i to be minimized. So this approach is not optimal. So there must be a better way.

Wait, perhaps the color can be assigned based on the parity and the smallest prime factor. For example:

- Even numbers: color 1.
- Odd numbers: if they are primes, color 2. If they are composite, color 1.

But this would work for the first example. Let's see:

For n=4:

a_2=1 (even).

a_3=2 (prime).

a_4=1 (even).

This matches the example.

For n=5:

a_5=2 (prime).

So for primes, color 2. Even numbers color 1. Composite odds color 1.

But then, 3 and 5 are primes. Their colors are 2. So they are coprimes but same color. Which is invalid. So this approach is wrong.

So the problem is that primes are coprimes with each other. So they must have different colors.

So how to assign colors to primes such that all primes have different colors. But that would require as many colors as the number of primes, which is not feasible for large n.

So this suggests that the minimal maximum color is equal to the maximum number of primes in any group of coprimes. For example, in the case of n=5, the primes 2,3,5 are all pairwise coprimes. So they form a clique of size 3. So the chromatic number is at least 3. So the maximum color must be at least 3.

But how to assign colors to other numbers.

For example, for even numbers (2,4,6, etc.), they are not coprime with each other. So they can all have color 1.

For primes (3,5,7, etc.), they are coprimes with each other. So each must have a unique color. But that's impossible for large n. So this approach is not feasible.

So there's a contradiction here. So perhaps the minimal maximum color is not the number of primes, but something else.

Alternatively, perhaps the color is determined by the parity of the number of prime factors. But I'm not sure.

Another idea: Let's think about the problem in terms of the parity of the numbers. For even numbers, assign color 1. For odd numbers, assign color based on whether they are primes or not. But how?

Wait, in the first example, the output is 1 2 1. So even numbers get color 1, primes get color 2. For n=4, the primes are 2 and 3. But a_2 is 1 (even), a_3 is 2 (prime). So the primes are not all assigned the same color. So perhaps the color of a prime is determined by something else.

Alternatively, perhaps the color is determined by the parity of the number's smallest prime factor. For example, even numbers (spf=2) get color 1. For primes, if the spf is 2 (even), color 1. For primes with spf>2 (which are odd primes), color 2. For composites, if their spf is even, color 1. If their spf is odd, color 2.

But this would lead to even numbers (spf=2) color 1, odd primes (spf is themselves) color 2, and odd composites (spf is odd) color 2. Then, two coprime numbers would have different colors. For example, 3 (spf 3, color 2) and 5 (spf 5, color 2). They are coprimes but same color. So this approach is invalid.

So this approach doesn't work.

Hmm. So perhaps the color is determined by the parity and the smallest prime factor. For example:

- Even numbers (spf=2) get color 1.
- Odd numbers: if they are primes, color 2. If they are composite, color 1.

But then, primes are all color 2, which is invalid for coprimes like 3 and 5.

So this approach is invalid.

Another idea: Let's think of the problem in terms of the parity of the number of prime factors. For example, numbers with even number of prime factors get color 1, odd get color 2. But this may not work.

Alternatively, think of the problem as a graph where nodes are numbers from 2 to n, and edges connect coprime pairs. Then, the problem is to find the minimal coloring of this graph.

The minimal coloring is the chromatic number. But how to compute this for large n?

But for large n, we need an efficient way to compute the colors.

Wait, but for the coprime graph, what's the chromatic number? For example, for numbers 2,3,5, which are all pairwise coprimes, they form a clique of size 3. So the chromatic number is at least 3. So for n >=5, the chromatic number is at least 3. But how to color the rest.

So perhaps the minimal maximum color is the maximum number of primes in any clique. But how to find that.

Alternatively, perhaps the chromatic number is 2 if there are no three primes, but 3 otherwise. But this is not necessarily true.

For example, if n=6, the primes are 2,3,5. So they form a clique of size 3. So the chromatic number is 3.

So for n >=5, the chromatic number is at least 3. But how to color the numbers.

An alternative approach: Assign colors based on the parity of the number, but for primes, assign a color that depends on their position in the list of primes.

But again, this would require many colors.

Alternatively, perhaps the color can be determined by the parity of the number's smallest prime factor. For example:

- Even numbers (spf=2) get color 1.
- Numbers with spf=3 get color 2.
- Numbers with spf=5 get color 3.
- etc.

This way, any two coprime numbers must have different spf's, so their colors are different. This approach ensures that all coprime pairs have different colors. But the maximum color is the number of distinct primes up to n. For n=1e5, this is about 10^4, which is way larger than the examples. So this approach is not optimal.

But the examples show that for n=4, the maximum color is 2. So this approach would not work for the examples.

Hmm. So there's a contradiction here. So perhaps there's a different approach.

Wait, let's look at the examples again.

In the first example, n=4:

a_2=1 (even), a_3=2 (prime), a_4=1 (even).

The primes are 2 and 3. But a_2 is 1 (even), which is the same as a_4. So even numbers can share the same color.

So perhaps the color is determined by the parity of the number, but for primes, their color is determined by something else.

Wait, but 2 is even and a prime. So in this case, it's assigned color 1. 3 is a prime and odd, assigned color 2.

So perhaps the color is 1 for even numbers, and for odd numbers, the color is 2 if the number is a prime, and 1 otherwise. But then, for n=5:

a_5=2 (prime, odd). So 3 and 5 are primes, both assigned color 2. But they are coprimes, so their colors must be different. So this approach is invalid.

So this approach doesn't work.

Another idea: The color is determined by the parity of the number of distinct prime factors. For example, even numbers (which have at least one prime factor, 2) have color 1. Odd numbers with even number of distinct prime factors (like 9=3^2) get color 1. Odd numbers with odd number of distinct prime factors (like 3, 5, 15=3*5) get color 2. But then, 3 and 5 are primes (odd number of factors), so color 2. But they are coprimes, so same color. Invalid.

So this approach is also invalid.

Hmm. So perhaps the key is to realize that the minimal maximum color is 2 when there are no three primes. But once there are three primes, the maximum color must be at least 3.

So for n >= 5, the maximum color is 3. But how to assign colors.

Let's think for n=5. The primes are 2,3,5. They form a clique of size 3, so need 3 colors. So a_2=1, a_3=2, a_5=3. Then, even numbers (4) get color 1. Composite odds (like 9, but n=5 is 5). So for 5, color 3.

But how to assign colors to other numbers.

For example, 6 is even, color 1. 7 is prime, color 2 (assuming the pattern). But 7 is coprime with 2,3,5. So a_7 must be different from 1,2,3. So color 4. But this would require more colors.

But this approach is not feasible for large n.

So perhaps there's a pattern where the color is determined by the parity and the number of times the number is a prime.

But I'm stuck. Let's think differently.

The problem requires that for any coprime pair (i,j), a_i != a_j. So the colors must be assigned such that all coprimes have different colors.

But what's the minimal number of colors needed?

The minimal number of colors is the chromatic number of the coprime graph. But what is this chromatic number?

In graph theory, the chromatic number of a graph is the smallest number of colors needed to color the vertices so that no two adjacent vertices share the same color.

The coprime graph is the complement of the non-coprime graph. So two numbers are connected if they are coprimes.

But what is the chromatic number of this graph?

For example, for n=5, the primes 2,3,5 form a clique of size 3. So the chromatic number is at least 3. And since other numbers are connected to these primes, their colors must be different from the primes' colors.

But how to color the other numbers.

For even numbers, they are not coprime with 2. So their color can be the same as 2's color. For example, even numbers can be color 1. Then, for primes 3 and 5, they need colors different from 1. But they are also coprimes with each other, so they need different colors. So 3 could be color 2, 5 color 3. Then, other numbers:

- 4 (even) color 1.
- 6 (even) color 1.
- 9 (composite, odd) color 1. Because 9 is coprime with 2 (no), 3 (yes). So a_9 must be different from 2's color (1). So color 2. But 9 is coprime with 5 (yes), so a_9 must be different from 5's color (3). So color 2 is allowed. So a_9=2.

But then, 9 is coprime with 4 (gcd 1). So a_9=2, a_4=1. Different, okay.

But 9 is coprime with 5 (a_9=2, a_5=3. Different. Okay.

So this approach works for 9.

But what about 15 (composite, odd). It is coprime with 2 (no), 3 (yes), 5 (yes). So a_15 must be different from 2's color (1), 3's color (2), 5's color (3). So a_15 must be color 4. Which increases the maximum color.

So this approach would require colors up to the number of primes plus something else, which is not feasible for large n.

So this suggests that the chromatic number is not bounded by a small number, but perhaps there's a way to assign colors with a maximum of 2 colors. But how?

Wait, but the examples show that for n=4, the maximum color is 2. For n=3, it's 2. For n=5, perhaps it's 3. So perhaps the maximum color is the number of primes in the set. But that's not possible for large n.

But the problem requires that the maximum a_i is minimized. So perhaps the minimal maximum is the maximum number of primes in any clique of coprimes. For example, if there's a clique of k primes, then the maximum color is at least k.

So the minimal maximum is the size of the largest clique of primes. Since primes are pairwise coprimes, the largest clique of primes is the number of primes in the set. But that's not possible for large n.

But this suggests that the problem's solution is to assign colors based on the parity and the smallest prime factor. For example:

- Even numbers: color 1.
- Odd numbers: if they are primes, assign colors in a way that they are all different. But this is not possible for large n.

But how to do this.

Alternatively, perhaps the color is determined by the parity of the number of times the number is a prime. For example, primes are assigned colors in a way that no two primes share the same color. But this requires a lot of colors.

But the examples show that for n=4, the primes are 2 and 3. a_2=1 (even), a_3=2 (prime). So the color of primes is determined by their order. But this approach would require as many colors as the number of primes, which is not feasible.

Hmm. I'm stuck. Let's look for a pattern in the examples.

In the first example, n=4:

a_2=1 (even), a_3=2 (prime), a_4=1 (even).

In the second example, n=3:

a_2=2 (even), a_3=1 (prime).

So the even numbers are assigned colors that are different from primes. But the primes are assigned colors in a way that they are different from each other.

So for n=3, primes are 2 and 3. a_2=2 (even), a_3=1 (prime). So the even primes (2) are assigned color 2, and odd primes (3) are assigned color 1.

But for n=4, a_2=1 (even), a_3=2 (prime). So even primes are assigned color 1, odd primes assigned color 2.

So perhaps the color of a prime is determined by its parity. Even primes (only 2) get color 1. Odd primes get color 2.

But then, for n=5, the primes are 2,3,5. So a_2=1 (even), a_3=2 (odd), a_5=2 (odd). But 3 and 5 are coprimes and have the same color. So this is invalid.

So this approach doesn't work.

Another idea: For even numbers, color 1. For primes, color 2 if the prime is 2, and color 3 if the prime is odd. But this would require more colors.

But for n=5, a_2=1 (even), a_3=3 (prime), a_5=3 (prime). But 3 and 5 are coprimes and have the same color. Invalid.

So this approach is not working.

Hmm. Let's think of the problem in terms of the parity of the numbers. For even numbers, color 1. For odd numbers, color 2 if they are primes, and color 1 otherwise. But this leads to primes having the same color, which is invalid.

But what if we assign colors to primes in a way that they are all different. For example, for primes, assign colors in the order they appear. But this would require as many colors as the number of primes, which is not feasible.

So perhaps the solution is to assign colors based on the parity of the number, but for primes, assign colors in a way that they are all different. But how to do this without increasing the maximum color.

Wait, but the examples show that for n=4, the maximum color is 2. So perhaps the solution is to assign colors as follows:

- Even numbers: color 1.
- Odd primes: color 2.
- Composite odds: color 1.

But this would fail for primes 3 and 5, which are coprimes and have the same color. So this is invalid.

So this approach is not correct.

Another idea: Let's think of the problem as follows. For any two coprime numbers, their colors must be different. So the color of a number x must be different from the colors of all numbers y that are coprime with x.

But how to assign colors in a way that this is satisfied with minimal maximum color.

An alternative approach is to assign colors based on the parity of the number of distinct prime factors of x. For example, even numbers (which have at least one prime factor) have color 1. Odd numbers with even number of distinct prime factors (like 9=3^2) have color 1. Odd numbers with odd number of distinct prime factors (like 3, 5, 15=3*5) have color 2. But then, primes (3,5) have color 2, which are coprimes, so same color. Invalid.

So this approach is not correct.

Hmm. I'm stuck. Let's look for another pattern.

In the examples, for n=4, the output is 1 2 1. For n=3, it's 2 1. So for even numbers, the color is 1 (n=4) or 2 (n=3). So there's no consistent pattern here.

Wait, but in the first example, a_2 is 1. In the second example, a_2 is 2. So perhaps the color of even numbers is determined by whether the number is a prime. For example, 2 is even and a prime. So for even primes, color is 1. For even composites, color is 1. For odd primes, color is 2. For odd composites, color is 1.

But this approach would assign a_2=1 (even prime), a_3=2 (odd prime), a_4=1 (even composite), a_5=2 (odd prime). But 3 and 5 are coprimes and have the same color. So invalid.

So this approach is not correct.

Another idea: The color is determined by the parity of the number, but for primes, the color is determined by their order. For example:

- Even numbers: color 1.
- Primes: color 2 if the prime is 2, color 3 if the prime is 3, color 4 if the prime is 5, etc.

But this would require as many colors as the number of primes, which is not feasible for large n.

But the examples show that for n=4, the maximum color is 2. So this approach is not correct.

Hmm. Perhaps the correct approach is to assign colors based on the parity of the number, but for primes, assign colors in a way that they are all different. But how to do this without increasing the maximum color.

Wait, perhaps the color of a number is 1 if it's even, and for odd numbers, the color is 2 if it's a prime, and 1 otherwise. But this would require that all primes are assigned color 2. But primes are coprimes with each other, so they need different colors. So this approach is invalid.

But what if the color of a prime is determined by the parity of the number of primes before it. For example, the first prime (2) is color 1, the second (3) color 2, the third (5) color 3, etc. But this would require as many colors as the number of primes, which is not feasible.

But for n=4, this would work. For n=3, the primes are 2 and 3. So a_2=1, a_3=2. Which matches the example.

For n=5, a_2=1, a_3=2, a_5=3. Even numbers (4) are color 1. Composite odds (like 9) are color 1. So for 9, which is coprime with 2 (no), 3 (yes), 5 (yes). So a_9 must be different from 2's color (1), 3's color (2), 5's color (3). So a_9 must be 4. Which increases the maximum color.

But this approach would require colors up to the number of primes, which is not feasible for large n.

But the problem's constraints allow n up to 1e5, and the maximum a_i is up to n. So this approach is possible, but the maximum color would be the number of primes up to n, which for 1e5 is about 10^4. But the examples show that for n=4, the maximum is 2, which is the number of primes. So perhaps this is the correct approach.

But how to implement this efficiently.

The steps would be:

1. Precompute the primes up to n using sieve.

2. Assign colors to primes in the order they appear. The first prime (2) gets color 1, the second (3) color 2, etc.

3. For non-primes, if even, color 1. If odd and composite, color 1.

But then, primes are assigned colors in order, and non-primes get color 1 or 2 (if odd composites).

Wait, but for non-primes:

- Even numbers: color 1.

- Odd composites: color 1.

So for example, 9 is an odd composite, color 1. But 9 is coprime with 3 (prime, color 2). So 9's color (1) is different from 3's color (2). So okay.

But 9 is also coprime with 5 (color 3). So 1 != 3. Okay.

So this approach works for primes and composites.

But what about two composites that are coprimes. For example, 9 and 25. Both are composite odds, color 1. They are coprimes, so their colors must be different. But they are the same. So this approach is invalid.

So this approach fails for composites that are coprimes.

So this suggests that the problem is more complex than initially thought.

Another idea: The color of a number x is determined by the parity of the number of distinct prime factors of x. For example:

- Even numbers: color 1.

- Odd numbers with even number of distinct prime factors: color 1.

- Odd numbers with odd number of distinct prime factors: color 2.

But then, primes (which have one distinct prime factor, odd) would be color 2. Composites like 9 (one distinct prime factor) would be color 2. So 9 and 3 are coprimes and have the same color. Invalid.

So this approach is invalid.

Hmm. I'm stuck. Let's think of the problem from another angle.

The problem requires that for any two coprimes, their colors are different. So the colors must form a proper coloring of the coprime graph.

The minimal maximum color is the chromatic number of the coprime graph.

But how to compute this for large n.

The coprime graph is the complement of the non-coprime graph. The non-coprime graph includes edges between numbers that share a common factor.

But the complement of the non-coprime graph is the coprime graph.

The chromatic number of the complement of a graph is not straightforward to compute.

But perhaps there's a pattern in the coprime graph's structure.

For example, the coprime graph for numbers 2 to n includes edges between all pairs of numbers that are coprime.

But this graph is not a complete graph. For example, 2 and 4 are not coprimes, so no edge.

But the primes form a clique in the coprime graph. So the chromatic number is at least the size of the largest clique of primes.

But how to color the rest.

Another idea: Assign colors based on the parity of the number, but for primes, assign colors in a way that they are all different. For example, even numbers get color 1. Odd primes get colors 2,3,4,... in order. Composite odds get color 1.

But this would require as many colors as the number of primes, which is not feasible for large n.

But the examples show that for n=4, the maximum color is 2. So this approach is not correct.

Wait, but for n=4, there are two primes (2 and 3). So a_2=1 (even), a_3=2 (prime). So the primes are assigned colors 1 and 2. So the first prime (2) is even, so color 1. The second prime (3) is odd, so color 2. Then, other primes (5) would be color 3, etc.

So for n=5, a_5=3. Then, even numbers get color 1. Composite odds get color 1. So 9 (composite, odd) gets color 1. But 9 is coprime with 3 (color 2), so okay. 9 is coprime with 5 (color 3), so okay. But 9 is coprime with 7 (color 4), which would be assigned to 7. So this approach would require colors up to the number of primes.

But for n=1e5, this would require 1e4 colors, which is possible but not optimal.

But the problem's examples show that the maximum color can be as low as 2. So perhaps there's a different approach.

Wait, perhaps the solution is to assign colors based on the parity of the number's smallest prime factor. For example:

- Even numbers (spf=2) get color 1.

- Numbers with spf=3 get color 2.

- Numbers with spf=5 get color 3.

- etc.

This way, any two coprime numbers will have different spf's, so different colors. This approach ensures that all coprimes have different colors. But the maximum color is the number of distinct primes up to n, which is about 1e4 for n=1e5. But this is not optimal. For example, in the first example, this approach would assign a_3=2 (spf=3), which is correct. a_2=1 (spf=2), a_4=1 (spf=2). So this matches the example. But for n=5, a_5=3 (spf=5). So the maximum is 3, which is higher than the example's 2. So this approach is not optimal.

But the problem requires the maximum a_i to be minimized. So this approach is not the optimal solution.

Hmm. So I'm stuck. Let's look for a pattern in the examples.

In the first example, the output is 1 2 1.

The primes are 2 and 3. a_2=1 (even), a_3=2 (prime). So the even prime is color 1, the odd prime is color 2.

In the second example, the output is 2 1. a_2=2 (even), a_3=1 (prime). So the even prime is color 2, the odd prime is color 1.

So the color of even primes (2) varies between examples. So there's no consistent pattern.

Another idea: The color of a number is determined by the number of distinct prime factors. For example:

- Even numbers: color 1.

- Odd numbers with one distinct prime factor (primes): color 2.

- Odd numbers with two or more distinct prime factors (composites): color 1.

But this approach would assign primes color 2. But primes are coprimes with each other, so they need different colors. So this approach is invalid.

Hmm. I'm stuck. Let's think about the problem's constraints. The maximum a_i is to be minimized. So the solution must use as few colors as possible.

The examples suggest that the solution uses colors based on parity for even numbers and primes. But how?

Wait, perhaps the color is determined by the parity of the number, but for primes, the color is determined by their parity. For example:

- Even numbers: color 1.

- Odd primes: color 2.

- Odd composites: color 1.

This way, even numbers and odd composites are color 1. Odd primes are color 2.

But this approach would assign a_3=2, a_5=2. But 3 and 5 are coprimes and have the same color. So this is invalid.

So this approach is invalid.

Another idea: The color is determined by the parity of the number, but for primes, the color is determined by their position in the list of primes. For example:

- Even numbers: color 1.

- Odd primes: color 2,3,4,... in order.

- Odd composites: color 1.

But this requires as many colors as the number of primes, which is not feasible for large n.

But the examples show that for n=4, the maximum color is 2. So this approach would work for the examples.

But for n=5, the primes are 2,3,5. So a_2=1 (even), a_3=2 (odd prime), a_5=3 (odd prime). So the maximum is 3. But how to assign colors to other numbers.

For example, 4 is even, color 1. 9 is composite, color 1. 9 is coprime with 3 (color 2) and 5 (color 3). So a_9=1 is different from 2 and 3. So okay.

But 9 is coprime with 4 (color 1). So a_9=1 and a_4=1. Same color, which is invalid.

So this approach is invalid.

Hmm. I'm stuck. Let's think of another way.

The problem requires that any two coprimes have different colors. So the color of a number x must be different from all numbers y that are coprimes with x.

But how to assign colors in a way that this is satisfied with minimal maximum color.

An alternative approach is to assign colors based on the parity of the number. But for primes, assign colors in a way that they are all different. But this requires as many colors as the number of primes.

But this is not feasible for large n.

But the examples show that for n=4, the maximum color is 2, which is the number of primes. So perhaps the solution is to assign colors to primes in a way that they are all different, and assign even numbers color 1, odd composites color 1.

But this would require colors up to the number of primes.

But how to implement this.

The steps would be:

1. Precompute all primes up to n using sieve.

2. For each prime, assign a unique color. The first prime (2) gets color 1, the second (3) color 2, etc.

3. For even numbers (non-primes), assign color 1.

4. For odd composites, assign color 1.

But this approach would work for the examples.

For n=4:

Primes are 2 and 3. a_2=1 (prime), a_3=2 (prime). Even numbers (4) get color 1. So output is 1 2 1. Which matches the example.

For n=3:

Primes are 2 and 3. a_2=1, a_3=2. Even numbers (none). Output is 1 2. But the example's output is 2 1. So this approach would give a different output but still valid.

But the problem allows any valid solution as long as the maximum is minimized.

So this approach would work.

But for n=5:

Primes are 2,3,5. a_2=1, a_3=2, a_5=3. Even numbers (4) get color 1. Odd composites (none). So output is 1 2 3 1. The maximum is 3.

But how to handle numbers like 9 (composite, odd) for n=9. a_9=1. But 9 is coprime with 3 (color 2) and 5 (color 3), so a_9=1 is different. So okay. But 9 is coprime with 7 (color 4), so a_9=1 is different. So okay.

But 9 is coprime with 4 (color 1). So a_9=1 and a_4=1. Same color. Which is invalid.

So this approach is invalid.

So this approach works for primes and even numbers, but fails for composite odds that are coprimes with even numbers.

So this approach is not correct.

Hmm. I'm stuck. Let's think of the problem differently.

The problem requires that any two coprimes have different colors. So the color of a number x must be different from all numbers y that are coprimes with x.

But how to assign colors in a way that this is satisfied with minimal maximum color.

An alternative approach is to assign colors based on the parity of the number. For even numbers, color 1. For odd numbers, color 2. But this approach fails for coprimes like 3 and 5 (both odd, color 2).

But what if we assign colors based on the parity of the number and the number of times it is a prime. For example, even numbers: color 1. Odd primes: color 2. Odd composites: color 1. This way, primes are all color 2, which is invalid.

But if we can find a way to assign colors to primes such that they are all different, but the maximum color is minimized.

But this requires as many colors as the number of primes, which is not feasible for large n.

But the examples show that for n=4, the maximum color is 2. So perhaps the solution is to assign colors to primes in a way that they are all different, but the maximum color is the number of primes.

But this would require that for each prime, the color is unique. But how to implement this efficiently.

But for n=1e5, this would require 1e4 colors, which is possible but not optimal.

But the problem requires the maximum a_i to be minimized. So this approach would not be optimal.

So I'm stuck. Let's look for a different approach.

Another idea: The color of a number is determined by the smallest prime factor. For even numbers (spf=2), color 1. For numbers with spf=3, color 2. For spf=5, color 3, etc. This way, any two coprimes have different spf's, so different colors. This approach ensures that all coprimes have different colors. The maximum color is the number of distinct primes up to n.

But this approach is not optimal. For example, in the first example, the maximum color is 2, which is the number of primes. So this approach works for the example.

But for n=5, the maximum color is 3, which is the number of primes. But the example's output for n=5 would require a maximum of 3.

But this approach is not optimal for larger n.

But the problem's examples show that the maximum color can be as low as 2, which is the number of primes for n=4.

So this approach is correct for the examples, but not optimal for larger n.

But the problem requires that the maximum a_i is minimized. So this approach may not be the optimal solution.

But given the time constraints, perhaps this is the best approach.

So the steps are:

1. Precompute the smallest prime factor (spf) for each number up to n.

2. For each number x from 2 to n:

   a. If x is even (spf=2), assign color 1.

   b. Else, assign color equal to the spf of x.

This way, coprimes have different spf's, so different colors. The maximum color is the number of distinct primes up to n.

But for n=4, the primes are 2 and 3. So the colors are 1 (for 2), 3 (for 3), and 1 (for 4). But the example's output is 1 2 1. So this approach would assign a_3=3, which is higher than the example's 2. So this approach is not correct.

So this approach is invalid.

Hmm. So what's the correct approach?

Let me think of the problem in terms of the examples. For n=4, the output is 1 2 1. The primes are 2 and 3. So a_2=1 (even), a_3=2 (prime), a_4=1 (even). So the color of the even prime (2) is 1, the color of the odd prime (3) is 2. So perhaps the color of a prime is determined by its parity. Even primes (only 2) get color 1. Odd primes get color 2. Composite odds get color 1.

But then, for n=5, the primes are 2,3,5. So a_2=1 (even), a_3=2 (odd), a_5=2 (odd). But 3 and 5 are coprimes and have the same color. So this approach is invalid.

So this approach is not correct.

Another idea: The color of a number is determined by the parity of the number. For even numbers, color 1. For odd numbers, color 2 if the number is a prime, and 1 otherwise. But this approach assigns all primes the same color (2), which is invalid for coprimes.

But if we can find a way to assign different colors to primes.

But how?

Perhaps the color of a prime is determined by the number of primes before it. For example, the first prime (2) is color 1, the second (3) is color 2, the third (5) is color 3, etc. But this requires as many colors as the number of primes, which is not feasible.

But for n=4, this would work. For n=5, a_5=3. Even numbers get color 1. Composite odds get color 1. So for 9 (composite), color 1. But 9 is coprime with 3 (color 2) and 5 (color 3), so okay. But 9 is coprime with 4 (color 1), which is invalid.

So this approach is invalid.

Hmm. I'm stuck. Let's think of the problem in terms of the required conditions.

The conditions are:

- For any coprime pair (i,j), a_i != a_j.

- The maximum a_i is minimized.

The key is to find a coloring where coprimes have different colors, and the maximum color is as small as possible.

The minimal maximum color is the chromatic number of the coprime graph.

But how to compute this for large n.

The coprime graph is the complement of the non-coprime graph. The non-coprime graph is a union of cliques for each prime.

For example, numbers divisible by 2 form a clique in the non-coprime graph. Similarly for 3, 5, etc.

The complement of this is the coprime graph, which is a graph where two numbers are connected if they are coprimes.

But what is the chromatic number of this graph?

For example, the primes form a clique in the coprime graph. So the chromatic number is at least the number of primes.

But this is not possible for large n. So there must be a different approach.

Wait, but the examples show that for n=4, the maximum color is 2. So the chromatic number is 2 for n=4.

But for n=5, the chromatic number is at least 3.

So perhaps the chromatic number is the maximum number of primes in any clique of coprimes. But how to find this.

But this seems difficult.

Another idea: The color of a number is determined by the parity of the number. For even numbers, color 1. For odd numbers, color 2. But this approach fails for coprimes like 3 and 5.

But what if we assign colors based on the parity and the smallest prime factor.

For example:

- Even numbers: color 1.

- Odd numbers: if their smallest prime factor is 3, color 2. If it's 5, color 3, etc.

But this approach would require as many colors as the number of primes, which is not feasible.

But for n=4, this approach would assign a_3=2, a_2=1, a_4=1. Which matches the example.

For n=5, a_5=3. So the maximum is 3.

But for n=9, a_9=2 (spf=3). So 9 is coprime with 5 (color 3), so okay. 9 is coprime with 4 (color 1), so okay. But 9 is coprime with 7 (color 4), so a_9=2 is different from 4. So okay.

But 9 and 15 (spf=3) would have color 2. They are coprimes (gcd(9,15)=3). So no edge. So colors can be same.

But 9 and 7 are coprimes. a_9=2, a_7=4. Different. So okay.

So this approach works for coprimes. But the maximum color is the number of primes.

But for n=1e5, this is not feasible.

But the problem's examples show that this approach works for small n, but the maximum color is the number of primes.

But the problem requires the maximum a_i to be minimized. So this approach is not optimal.

But perhaps there's a way to assign colors with a maximum of 2 colors.

But how?

Wait, let's think of the problem in terms of the following:

If two numbers are coprimes, then one is even and the other is odd. Because if both are even, they share 2 as a common factor. So they are not coprimes. So any coprime pair must consist of one even and one odd.

So the parity of the numbers determines their colors. For example, even numbers get color 1, odd numbers get color 2. This way, any coprime pair has different colors. Because one is even, the other is odd.

But this approach works for all coprime pairs. Because any coprime pair must consist of one even and one odd number.

But what about 3 and 5? Both are odd, so they are coprimes. But according to this approach, they would have the same color (2), which is invalid.

So this approach is invalid.

Ah, here's the mistake. 3 and 5 are both odd and coprimes. So their colors must be different. So this approach fails.

So this approach is not correct.

But how to handle this case.

The problem is that two odd numbers can be coprimes. So their colors must be different.

So the parity approach is not sufficient.

So the solution must assign colors in a way that not only even and odd are different, but also that any two odd coprimes have different colors.

So this suggests that the color of a number must be determined by more than just its parity.

So perhaps the color is determined by the parity of the number and the parity of the number of distinct prime factors.

But this is getting complicated.

Another idea: For even numbers, color 1. For odd numbers, assign colors based on their parity in the list of primes. For example:

- Even numbers: color 1.

- Odd primes: color 2,3,4,... in order.

- Odd composites: color 1.

But this approach requires as many colors as the number of primes. For example, for n=5, the primes are 2,3,5. a_2=1 (even), a_3=2 (prime), a_5=3 (prime). Even numbers (4) get color 1. Composite odds (9) get color 1. So 9 is coprime with 3 (color 2) and 5 (color 3), so okay. But 9 is coprime with 4 (color 1), which is invalid.

So this approach is invalid.

Hmm. I'm stuck. Let's think of the problem from the perspective of the required conditions.

The conditions are:

1. For any coprime pair (i,j), a_i != a_j.

2. The maximum a_i is minimized.

So the solution must assign colors such that all coprimes have different colors, and the maximum color is as small as possible.

The minimal maximum color is the chromatic number of the coprime graph.

But how to compute this for large n.

The coprime graph is a graph where two numbers are connected if they are coprimes.

The chromatic number of this graph is the minimal number of colors needed.

But how to find this.

The coprime graph is the complement of the non-coprime graph. The non-coprime graph is a union of cliques for each prime.

For example, numbers divisible by 2 form a clique in the non-coprime graph. Similarly for 3, 5, etc.

The complement of this is the coprime graph.

The chromatic number of the coprime graph is the minimal number of colors needed to color the nodes such that no two coprimes have the same color.

But how to find this.

The chromatic number of the coprime graph is at least the maximum size of a clique in the coprime graph. A clique in the coprime graph is a set of numbers where every pair is coprimes. The largest such clique is the set of primes.

But the size of this clique is the number of primes in the set. So the chromatic number is at least the number of primes.

But for n=4, the number of primes is 2, and the chromatic number is 2. For n=5, the number of primes is 3, and the chromatic number is 3.

So the chromatic number is equal to the number of primes.

But how to assign colors to the nodes.

For the primes, each must have a unique color. For non-primes, their colors must be different from the colors of all primes that are coprimes with them.

But how to assign colors to non-primes.

For even numbers (non-primes), they are not coprimes with 2. So their color can be the same as 2's color.

For odd composites, they are coprimes with all primes that are not their factors. So their color must be different from all primes that are coprimes with them.

But this seems complex.

But perhaps the solution is to assign colors to primes in a way that they are all different, and assign non-primes colors that are different from the colors of all primes that are coprimes with them.

But how to do this efficiently.

An efficient way is to assign colors to primes in a way that they are all different. For example, assign each prime a unique color. For non-primes, assign a color that is the same as one of the primes that shares a common factor with them.

For example:

- Even numbers: color 1 (same as prime 2).

- Odd composites: color 2 (same as prime 3 if they share 3 as a factor), 3 (same as prime 5 if they share 5 as a factor), etc.

But this approach ensures that non-primes have colors that are the same as primes that share a common factor with them. So coprimes with non-primes will have different colors.

But this approach may not work for all cases.

For example, consider the number 15 (3*5). It is coprime with 2 (color 1), 7 (color 3), etc. So 15's color must be different from 1 and 3. So if 15 is assigned color 2 (same as 3's color), then 15 and 3 are not coprimes (gcd 3), so they can have the same color. But 15 and 7 are coprimes, so 15's color must be different from 7's color (3). So 15's color is 2, which is different from 3. So this is okay.

So this approach works.

So the steps are:

1. Assign each prime a unique color. The first prime (2) gets color 1, the second (3) gets color 2, the third (5) gets color 3, etc.

2. For non-primes:

   a. If the number is even, assign color 1 (same as 2's color).

   b. If the number is odd and composite, assign color equal to the smallest prime factor's color.

This way, any coprime pair will have different colors.

For example:

- 3 (prime) is color 2.

- 5 (prime) is color 3.

- 15 (composite, spf 3) is color 2.

- 15 and 5 are not coprimes (gcd 5), so can have same color.

- 15 and 7 (prime) are coprimes. 15's color is 2, 7's color is 4. So different.

- 15 and 2 are coprimes (gcd 1). 15's color is 2, 2's color is 1. Different.

So this approach works.

But what about two composites that are coprimes. For example, 9 (spf 3, color 2) and 25 (spf 5, color 3). They are coprimes. Their colors are 2 and 3. Different. So okay.

Another example: 9 and 7 (spf 7, color 4). They are coprimes. Colors 2 and 4. Different.

So this approach works.

So the steps are:

- For each number x from 2 to n:

   if x is a prime, assign a unique color in the order of primes.

   else:

      if x is even, assign color 1.

      else, assign color equal to the color of the smallest prime factor of x.

This approach ensures that:

- For coprime pairs (i,j):

   if one is even and the other is odd, their colors are different.

   if both are odd, then they are coprimes only if they are not sharing any prime factors. So their colors are based on their smallest prime factors. Since they are coprimes, their smallest prime factors must be different. So their colors are different.

So this approach ensures that all coprime pairs have different colors.

The maximum color is the number of primes up to n.

But this is not optimal. For example, in the first example, the maximum color is 2 (primes 2 and 3). Which matches the example.

In the second example, the primes are 2 and 3. a_2=1, a_3=2. Even numbers (none). So output is 1 2. But the example's output is 2 1. So this approach gives a different output but is still valid.

So this approach is correct.

But for n=5, the primes are 2,3,5. a_2=1, a_3=2, a_5=3. Even numbers (4) get color 1. Composite odds (none). So output is 1 2 3 1. The maximum is 3.

But how to implement this.

The steps are:

1. Generate all primes up to n using sieve.

2. Assign each prime a unique color in the order they appear.

3. For each number x from 2 to n:

   if x is a prime, assign its color.

   else:

      if x is even, assign color 1.

      else, assign the color of the smallest prime factor of x.

This way, the maximum color is the number of primes.

But for n=1e5, this would require 1e4 colors, which is possible but may not be optimal.

But the problem's examples show that the maximum color can be as low as 2. So this approach is not optimal.

But perhaps this is the correct approach, and the examples are just special cases.

But the problem requires that the maximum a_i is minimized. So this approach may not be optimal.

But given the time constraints, perhaps this is the best approach.

So the code steps are:

- Precompute the smallest prime factor (spf) for each number up to n.

- Generate the list of primes in order.

- For each number x from 2 to n:

   if x is a prime, assign color based on its position in the primes list.

   else:

      if x is even, color 1.

      else, color of spf[x].

This way, the maximum color is the number of primes.

But for n=4, the primes are 2 and 3. So a_2=1, a_3=2, a_4=1. Which matches the example.

For n=3, primes are 2 and 3. a_2=1, a_3=2. Output is 1 2. But the example's output is 2 1. So this approach is correct.

But the example's output is different but valid.

So this approach is correct.

Now, let's implement this.

The code steps:

1. Read n.

2. Compute spf for each number up to n.

3. Generate the list of primes in order.

4. For each number x from 2 to n:

   if x is a prime, assign color based on its index in the primes list (starting from 1).

   else:

      if x is even, color 1.

      else, color is spf[x]'s color.

But how to get the color of spf[x].

We can create a dictionary or array to store the color of each number.

But since spf[x] is a prime, we can precompute the color for each prime and then use that.

So the code:

Compute spf for each x.

Generate primes list.

Create a color array.

For each prime in primes list, assign color in order.

For non-primes:

   if even, color 1.

   else, color is the color of spf[x].

So the code:

n = int(input())

if n < 2:
    print()
    exit()

# Compute smallest prime factors
spf = list(range(n+1))
for i in range(2, int(n**0.5)+1):
    if spf[i] == i:
        for j in range(i*i, n+1, i):
            if spf[j] == j:
                spf[j] = i

# Generate primes list
primes = []
for i in range(2, n+1):
    if spf[i] == i:
        primes.append(i)

# Assign colors
color = [0]*(n+1)
for idx, p in enumerate(primes):
    color[p] = idx + 1

# Process each number from 2 to n
output = []
for x in range(2, n+1):
    if spf[x] == x:
        # prime
        output.append(str(color[x]))
    else:
        if x % 2 == 0:
            output.append('1')
        else:
            output.append(str(color[spf[x]]))

print(' '.join(output))

Testing this code for the examples.

Example 1:

n=4.

Primes are 2,3.

color[2]=1, color[3]=2.

For x=2: prime, color 1.

x=3: prime, color 2.

x=4: even, color 1.

Output: 1 2 1. Which matches the example.

Example 2:

n=3.

Primes are 2,3.

color[2]=1, color[3]=2.

x=2: prime, color 1.

x=3: prime, color 2.

Output: 1 2. But the example's output is 2 1. So this approach is correct, but the example's output is different but valid.

But the problem allows any valid solution. So this code is correct.

Another test case: n=5.

Primes are 2,3,5.

color[2]=1, color[3]=2, color[5]=3.

x=2: 1.

x=3: 2.

x=4: even, 1.

x=5:3.

Output: 1 2 1 3.

But the maximum is 3.

But according to the problem's requirements, this is correct.

But what about x=9 (n=9):

x=9: spf=3. color[3] is 2. So color is 2.

But 9 is coprime with 5 (color 3). So 2 != 3. Okay.

9 is coprime with 7 (color 4). 2 !=4. Okay.

So this approach works.

Thus, this code should be correct.

Analysis complete. Proceeding with the solution.
</think>