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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  long long int gcd(long long int a, long long int b) {
    if (!b)
      return a;
    else
      return gcd(b, a % b);
  }
  const int MAXN = 1e6 + 1;
  int prime[MAXN];
  long long int h[MAXN], g[MAXN];
  vector<int> p;
  mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    prime[1] = 1;
    for (int i = 2; i <= n; i++) {
      if (!prime[i]) {
        prime[i] = i;
        p.push_back(i);
      }
      for (int x : p) {
        if (x * i > n) break;
        prime[x * i] = x;
        if (x == prime[i]) break;
      }
    }
    for (int x : p)
      h[x] = uniform_int_distribution<long long int>(0, 9223372036854775807)(rng);
    long long int ans = 0, cur = 0;
    map<long long int, int> gg;
    for (int i = 2; i <= n; i++) {
      int j = i;
      while (j != 1) {
        int cnt = 0;
        int pp = prime[j];
        while (j % pp == 0) {
          j /= pp;
          cnt++;
        }
        if (cnt % 2) cur = cur ^ h[pp];
      }
      g[i] = cur;
      ans = ans ^ g[i];
      gg[cur] = i;
    }
    if (!ans) {
      cout << n << '\n';
      for (int i = 1; i <= n; i++) cout << i << " ";
    } else {
      for (int i = 1; i <= n; i++) {
        if ((g[i] ^ ans) == 0) {
          cout << n - 1 << '\n';
          for (int j = 1; j <= n; j++) {
            if (i == j) continue;
            cout << j << " ";
          }
          return 0;
        }
      }
      for (int i = 2; i <= n; i++) {
        if (gg[ans ^ g[i]]) {
          cout << n - 2 << '\n';
          int x = gg[ans ^ g[i]];
          for (int j = 1; j <= n; j++) {
            if (j == i || j == x) continue;
            cout << j << " ";
          }
          return 0;
        }
      }
      cout << n - 3 << '\n';
      for (int i = 1; i < n; i++) {
        if (i == 2 || i == n / 2) continue;
        cout << i << " ";
      }
    }
  }
  