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 n, ans;
  mt19937_64 rng(1919810);
  unsigned long long h[1000005];
  bitset<1000005> bs, pri;
  unordered_map<unsigned long long, int> mp;
  int main() {
    scanf("%d", &n), ans = n;
    for (int i = 2; i <= n; i++)
      if (!pri[i]) {
        unsigned long long H = rng();
        unsigned long long d = 1;
        for (;;) {
          d *= i;
          if (d > n) break;
          for (int j = d; j <= n; j += d) h[j] ^= H, pri[j] = 1;
        }
      }
    for (int i = 2; i <= n; i++) h[i] ^= h[i - 1];
    unsigned long long H = 0;
    for (int i = 1; i <= n; i++) H ^= h[i];
    if (H == 0)
      ans = n;
    else {
      bool ok = 0;
      ans--;
      for (int i = 1; i <= n; i++)
        if (h[i] == H) {
          bs[i] = 1;
          ok = 1;
          break;
        }
      if (!ok) {
        bool ok = 0;
        ans--;
        for (int i = 1; i <= n; i++) {
          if (mp.find(h[i]) != mp.end()) {
            bs[mp[h[i]]] = 1;
            bs[i] = 1;
            ok = 1;
            break;
          }
          mp[H ^ h[i]] = i;
        }
        if (!ok) {
          ans--;
          bs[2] = 1;
          bs[(n - 1) / 2] = 1;
          bs[n] = 1;
        }
      }
    }
    printf("%d\n", ans);
    for (int i = 1; i <= n; i++)
      if (bs[i] == 0) printf("%d ", i);
  }
  