1. Concise Problem Statement  
Given an integer N (1 ≤ N ≤ 2^20), assign to 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 the number of colors M used. Output M on the first line and any valid coloring c[1..N] on the second line.

2. Detailed Editorial  

  A. Reformulation as a graph‐coloring problem  
  Define a graph G with vertices {1,2,…,N} and an edge between any two numbers A, B if one divides the other. We seek a proper vertex coloring of G using the fewest colors.  

  B. Perfectness and chain‐decomposition  
  The divisibility relation is a partial order; its comparability graph is a perfect graph. By Dilworth’s theorem (and properties of perfect graphs), the chromatic number equals the size of the largest clique in G, which for a comparability graph is the length of the longest chain under divisibility.  

  C. Characterizing the longest divisor‐chain  
  A “divisor chain” is a sequence 1 = d_0 < d_1 < … < d_k ≤ N where each d_{i+1} is a multiple of d_i. To maximize k+1, you would at each step multiply by the smallest prime possible—so the longest chain up to N is  
    1 → 2 → 2^2 → … → 2^k   with 2^k ≤ N < 2^{k+1}.  
  Hence the minimum number of colors M = k+1 = ⌊log₂N⌋ + 1.  

  D. Constructing an optimal coloring  
  It suffices to assign to each i the length of the longest divisor‐chain ending at i (this is a standard DP/mex‐based greedy coloring):  
    dp[i] = 1 + max{ dp[d] : d | i, d < i }.  
  Then dp[1] = 1, and finally M = max_i dp[i], with c[i] = dp[i].  

  E. Efficient implementation  
  1. Sieve‐like DP in O(N log N):  
     - Initialize dp[i] = 1 for all i.  
     - For i from 1 to N: for each multiple j = 2i, 3i, … ≤ N, do dp[j] = max(dp[j], dp[i] + 1).  
  2. Or, even faster in O(N log log N), compute the total number of prime factors (with multiplicity) Ω(i) via a smallest‐prime‐factor sieve, then observe dp[i] = Ω(i) + 1.  

  Both approaches produce an optimal M and a valid coloring in time well within the limits.  

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

static const int MAXN = (1 << 20);  // upper bound on N

int n;
int answer[MAXN];           // answer[i] will hold the color of i
vector<int> adj[MAXN];      // adj[i] will store all proper divisors of i
bool used[MAXN];            // temporary marker array for mex

void read_input() {
    cin >> n;
}

void solve() {
    // 1) Initialize every number with color = 1
    for (int i = 1; i <= n; i++) {
        answer[i] = 1;
    }
    // 2) Build the list of proper divisors for each i
    //    For each i, we append i to the adj-list of all multiples j = 2*i,3*i,...
    for (int i = 1; i <= n; i++) {
        for (int j = i * 2; j <= n; j += i) {
            adj[j].push_back(i);
        }
    }
    // 3) Greedy coloring in increasing order (mex of colors of divisors)
    for (int i = 1; i <= n; i++) {
        // Mark colors used by all proper divisors of i
        for (int d : adj[i]) {
            used[ answer[d] ] = true;
        }
        // Find smallest color ≥1 not marked
        while (used[ answer[i] ]) {
            answer[i]++;
        }
        // Clear markers for next i
        for (int d : adj[i]) {
            used[ answer[d] ] = false;
        }
    }
    // 4) Find the maximum color used
    int max_color = 1;
    for (int i = 1; i <= n; i++) {
        max_color = max(max_color, answer[i]);
    }
    // 5) Output
    cout << max_color << "\n";
    for (int i = 1; i <= n; i++) {
        cout << answer[i] << (i < n ? ' ' : '\n');
    }
}

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

    read_input();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys
sys.setrecursionlimit(10**7)

def main():
    data = sys.stdin.read().strip().split()
    n = int(data[0])

    # We'll compute Ω[i] = number of prime factors of i (with multiplicity)
    # using a sieve for the smallest prime factor (spf).
    spf = [0] * (n+1)
    Omega = [0] * (n+1)
    # Standard sieve to fill spf[i] = smallest prime dividing i
    for i in range(2, n+1):
        if spf[i] == 0:  # i is prime
            for j in range(i, n+1, i):
                if spf[j] == 0:
                    spf[j] = i
    # Now compute Omega[i] = Omega[i//spf[i]] + 1
    for i in range(2, n+1):
        Omega[i] = Omega[i // spf[i]] + 1
    # Our coloring dp[i] = Omega[i] + 1
    dp = [Omega[i] + 1 for i in range(n+1)]
    # The number of colors needed is max dp[i]
    M = max(dp[1:])
    # Output
    out = []
    out.append(str(M))
    out.append(" ".join(str(dp[i]) for i in range(1, n+1)))
    print("\n".join(out))

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

5. Compressed Editorial  
Use the fact that the divisibility graph on {1..N} is perfect, so its chromatic number equals the size of its largest chain under divisibility. The maximum chain is 1→2→4→…→2^k with 2^k≤N, so M=⌊log₂N⌋+1. Color each i by the length of the longest chain ending at i, computable either via a sieve‐style DP in O(N log N) or by computing Ω(i) (total prime‐factor count) in O(N log log N), since dp[i]=Ω(i)+1.