<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

355. Numbers Painting
Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard



Dr. Vasechkin wants to paint all numbers from 1 to N in such a way that if number A is divisible by number B, numbers A and B have different colors.

Help Dr. Vasechkin to find such a painting, where the number of the colors used is minimal.

Input
The input contains integer number N ().

Output
Write the number of the colors M in the desired painting in the first line of the output. In the second line of the output write the desired painting of numbers from 1 to N. The used colors should be represented by numbers from 1 to M. If there are several solutions, choose any of them.

Example(s)
sample input
sample output
12
4
1 2 2 3 2 3 2 4 3 3 2 4

<|response|>
1. Abridged Problem Statement  
Given an integer N (1 ≤ N ≤ 2^20), assign each integer i in [1..N] a color c[i] (an integer in [1..M]) so that whenever A is divisible by B (A ≠ B), c[A] ≠ c[B]. Minimize M and output any valid coloring.

2. Key Observations  
- We form a graph on vertices {1..N} with edges connecting any two numbers when one divides the other.  
- This graph is the comparability graph of the divisibility partial order, which is a perfect graph.  
- In any perfect graph, the minimum number of colors (chromatic number) equals the size of the largest clique.  
- A clique in this graph corresponds to a chain under divisibility: d₀|d₁|…|d_k.  
- The longest chain up to N is 1 → 2 → 2² → … → 2^k, where 2^k ≤ N < 2^(k+1).  
- Hence the minimum number of colors needed is M = k+1 = ⌊log₂ N⌋ + 1.  
- One valid coloring is to assign to each i the length of the longest divisor‐chain ending at i. It can be shown that this is exactly Ω(i)+1, where Ω(i) is the total number of prime factors of i counting multiplicity.

3. Full Solution Approach  
a. Compute an array spf[1..N], where spf[x] = smallest prime factor of x.  
   - Initialize spf[x] = 0 for all x.  
   - For each p from 2 to N, if spf[p] == 0 (meaning p is prime), then for j = p, 2p, 3p, … ≤ N, set spf[j] = p if it’s still 0.  
b. Compute Ω[i] for i = 1..N:  
   - Ω[1] = 0.  
   - For i ≥ 2, Ω[i] = Ω[i / spf[i]] + 1.  
c. Define c[1..N] by c[i] = Ω[i] + 1.  
   - This ensures that along any divisor‐chain, colors strictly increase, so no two divisible numbers share color.  
d. The number of colors used is M = max_{1≤i≤N} c[i], which equals ⌊log₂ N⌋ + 1.  
e. Output M, then the sequence c[1], c[2], …, c[N].

Time complexity: O(N log log N) for the sieve and O(N) for the Ω computation and output.  
Memory: O(N).

4. C++ Implementation with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N;
    cin >> N;

    // 1) Compute smallest prime factor (spf) for each number up to N.
    vector<int> spf(N+1, 0);
    for (int p = 2; p <= N; ++p) {
        if (spf[p] == 0) {
            // p is prime
            for (int j = p; j <= N; j += p) {
                if (spf[j] == 0) {
                    spf[j] = p;
                }
            }
        }
    }

    // 2) Compute Omega(i): total number of prime factors of i (with multiplicity).
    vector<int> Omega(N+1, 0);
    // Omega[1] = 0 by definition
    for (int i = 2; i <= N; ++i) {
        int smallest = spf[i];
        int reduced = i / smallest;
        Omega[i] = Omega[reduced] + 1;
    }

    // 3) Define coloring c[i] = Omega[i] + 1
    vector<int> color(N+1);
    int M = 0;
    for (int i = 1; i <= N; ++i) {
        color[i] = Omega[i] + 1;
        M = max(M, color[i]);
    }

    // 4) Output result
    cout << M << "\n";
    for (int i = 1; i <= N; ++i) {
        cout << color[i] << (i < N ? ' ' : '\n');
    }

    return 0;
}
```

5. Python Implementation with Detailed Comments  
```python
import sys
def main():
    data = sys.stdin.read().split()
    N = int(data[0])

    # 1) Smallest prime factor sieve
    spf = [0] * (N+1)
    for p in range(2, N+1):
        if spf[p] == 0:  # p is prime
            for j in range(p, N+1, p):
                if spf[j] == 0:
                    spf[j] = p

    # 2) Compute Omega[i] = number of prime factors (with multiplicity)
    Omega = [0] * (N+1)
    # Omega[1] = 0
    for i in range(2, N+1):
        Omega[i] = Omega[i // spf[i]] + 1

    # 3) Build coloring: c[i] = Omega[i] + 1
    colors = [str(Omega[i] + 1) for i in range(1, N+1)]
    M = max(int(c) for c in colors)

    # 4) Output
    sys.stdout.write(str(M) + "\n")
    sys.stdout.write(" ".join(colors) + "\n")

if __name__ == "__main__":
    main()
```