<|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
The implementation uses the equivalent greedy graph-coloring directly on the divisibility graph:
a. For every i from 1 to N, build adj[i] = the list of proper divisors of i. This is done with a sieve-style loop: for each i, append i to adj[j] for all multiples j = 2i, 3i, … ≤ N.
b. Initialize answer[i] = 1 for all i.
c. Process i from 1 to N in increasing order. Mark the colors of all already-colored divisors of i (a `used` array), then advance answer[i] to the smallest color not marked (the mex), and clear the markers.
   - Because divisors are processed before their multiples, this greedy assigns each i the smallest color not used by any of its divisors, giving a proper coloring.
d. The number of colors used is M = max_{1≤i≤N} answer[i], which equals ⌊log₂ N⌋ + 1 (one plus the maximum number of prime factors with multiplicity, i.e. the longest divisor chain).
e. Output M, then the sequence answer[1], answer[2], …, answer[N].

Time complexity: O(N log N) to build the divisor lists and run the greedy coloring.
Memory: O(N log N) for the divisor lists (or O(N) if Ω is used instead, as in the Python version below).

4. C++ Implementation
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;

void read() { cin >> n; }

void solve() {
    // Build the divisibility DAG: every proper divisor of i is a neighbour of
    // i. A valid painting needs different colors for any A divisible by B, so
    // the chromatic requirement is exactly a graph coloring on this DAG.
    //
    // Process numbers in increasing order and greedily assign each i the
    // smallest color not used by its (already colored) divisors. The number of
    // colors equals the length of the longest divisor chain, i.e. one plus the
    // number of prime factors counted with multiplicity, which is optimal.

    vector<int> answer(n + 1, 1);
    vector<vector<int>> adj(n + 1);
    for(int i = 1; i <= n; i++) {
        for(int j = i * 2; j <= n; j += i) {
            adj[j].push_back(i);
        }
    }

    vector<char> used(n + 2, 0);
    for(int i = 1; i <= n; i++) {
        for(int v: adj[i]) {
            used[answer[v]] = 1;
        }
        while(used[answer[i]]) {
            answer[i]++;
        }
        for(int v: adj[i]) {
            used[answer[v]] = 0;
        }
    }

    int mx_col = 1;
    for(int i = 1; i <= n; i++) {
        mx_col = max(mx_col, answer[i]);
    }

    cout << mx_col << '\n';
    for(int i = 1; i <= n; i++) {
        cout << answer[i] << " ";
    }
    cout << '\n';
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    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()
```