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;
  mt19937_64 rnd(time(0));
  int n, f[1000005], lst[1000005];
  unsigned long long num[1000005], p[1000005];
  bool out[1000005];
  void print() {
    int cnt = 0;
    for (int i = 1; i <= n; i++)
      if (!out[i]) cnt++;
    cout << cnt << '\n';
    for (int i = 1; i <= n; i++)
      if (!out[i]) cout << i << ' ';
    exit(0);
  }
  unordered_map<unsigned long long, int> mp;
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    for (int i = 2; i <= n; i++) f[i] = 1;
    lst[1] = 1;
    for (int i = 2; i <= n; i++)
      if (f[i]) {
        for (int j = i; j <= n; j += i) lst[j] = i, f[j] = 0;
        num[i] = rnd();
      }
    for (int i = 1; i <= n; i++) {
      if (!num[i]) {
        int x = i;
        while (x != 1) {
          num[i] ^= num[lst[x]];
          x /= lst[x];
        }
      }
      p[i] = p[i - 1] ^ num[i];
    }
    unsigned long long now = 0;
    for (int i = 1; i <= n; i++) now = now ^ p[i];
    if (!now) print();
    for (int i = 1; i <= n; i++)
      if (now == p[i]) {
        out[i] = 1;
        print();
      }
    for (int i = 1; i <= n; i++) mp[p[i]] = i;
    for (int i = 1; i <= n; i++)
      if (mp.count(now ^ p[i])) {
        out[mp[now ^ p[i]]] = 1;
        out[i] = 1;
        print();
      }
    out[n / 2] = 1;
    out[n] = 1;
    out[2] = 1;
    print();
    return 0;
  }
  