description: Let's call a set of positive integers a_1, a_2, ..., a_k quadratic if the product of the factorials of its elements is a square of an integer, i. e. ∏_{i=1}^{k} a_i! = m^2, for some integer m.
  
  You are given a positive integer n.
  
  Your task is to find a quadratic subset of a set 1, 2, ..., n of maximum size. If there are multiple answers, print any of them.
  
  Input
  
  A single line contains a single integer n (1 ≤ n ≤ 10^6).
  
  Output
  
  In the first line, print a single integer — the size of the maximum subset. In the second line, print the subset itself in an arbitrary order.
  
  Examples
  
  Input
  
  
  1
  
  
  Output
  
  
  1
  1 
  
  
  Input
  
  
  4
  
  
  Output
  
  
  3
  1 3 4 
  
  
  Input
  
  
  7
  
  
  Output
  
  
  4
  1 4 5 6 
  
  
  Input
  
  
  9
  
  
  Output
  
  
  7
  1 2 4 5 6 7 9 
solution: #include <bits/stdc++.h>
  using namespace std;
  int a[1000001];
  long long h[1000001];
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  uniform_int_distribution<long long> uid = uniform_int_distribution<long long>();
  void PrintAns(int n, int x, int y) {
    int c = n - (x != 0) - (y != 0);
    printf("%d\n", c);
    for (int i = 1; i <= n; ++i) {
      if (i == x || i == y) {
        continue;
      }
      printf("%d ", i);
    }
    printf("\n");
  }
  void Find(int n) {
    long long H = 0;
    for (int i = 1; i <= n; ++i) {
      H ^= h[i];
    }
    if (H == 0) {
      PrintAns(n, 0, 0);
      return;
    }
    for (int i = 1; i <= n; ++i) {
      if (h[i] == H) {
        PrintAns(n, i, 0);
        return;
      }
    }
    map<long long, int> M;
    for (int i = 2; i <= n; ++i) {
      auto it = M.find(H ^ h[i]);
      if (it != M.end()) {
        PrintAns(n, it->second, i);
        return;
      }
      M[h[i]] = i;
    }
    M.clear();
    Find(n - 1);
  }
  int main() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      a[i] = i;
    }
    for (int i = 2; i <= n; ++i) {
      if (a[i] != i) {
        continue;
      }
      long long hi = uid(rng);
      for (int j = i; j <= n; j += i) {
        int x = a[j];
        long long hj = h[j];
        for (; x % i == 0; x /= i) {
          hj ^= hi;
        }
        a[j] = x, h[j] = hj;
      }
    }
    for (int i = 2; i <= n; ++i) {
      h[i] ^= h[i - 1];
    }
    Find(n);
    return 0;
  }
  