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;
  const int N = 1e6 + 2;
  unsigned long long h[N];
  int p[N];
  bool ed[N], flg[N];
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
    unordered_map<unsigned long long, int> mp;
    int n, m = 0, i, j, k;
    unsigned long long tot = 0;
    cin >> n;
    for (i = 2; i <= n; i++) {
      if (!ed[i]) p[m++] = i, h[i] = rnd();
      for (j = 0; j < m && (k = i * p[j]) <= n; j++) {
        ed[k] = 1;
        h[k] = h[i] ^ h[p[j]];
        if (i % p[j] == 0) break;
      }
      if (1 ^ n - i & 1) tot ^= h[i];
    }
    for (i = 3; i <= n; i++) h[i] ^= h[i - 1];
    fill(flg + 1, flg + n + 1, 1);
    if (tot == 0) goto out;
    for (i = 1; i <= n; i++)
      if (tot == h[i]) {
        flg[i] = 0;
        goto out;
      }
    for (i = 1; i <= n; i++) {
      auto it = mp.find(tot ^ h[i]);
      if (it != mp.end()) {
        flg[i] = flg[it->second] = 0;
        goto out;
      }
      mp[h[i]] = i;
    }
    flg[n >> 1] = flg[2] = flg[n] = 0;
  out:
    cout << accumulate(flg + 1, flg + n + 1, 0) << '\n';
    for (int i = 1; i <= n; i++)
      if (flg[i]) cout << i << ' ';
  }
  