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 c = 1000005;
  int n, pr[c];
  unsigned long long ert[c], xo;
  void co(vector<int> sz) {
    int si = sz.size();
    cout << n - si << "\n";
    for (int i = 1; i <= n; i++) {
      bool baj = 0;
      for (auto x : sz)
        if (x == i) baj = 1;
      if (!baj) cout << i << " ";
    }
    cout << "\n";
    exit(0);
  }
  void solve() {
    if (!xo) co({});
    map<unsigned long long, int> m;
    for (int i = 1; i <= n; i++) {
      if (xo == ert[i]) co({i});
      m[ert[i]] = i;
    }
    for (auto x : m) {
      unsigned long long p = xo ^ x.first;
      if (m.find(p) != m.end()) {
        co({x.second, m[p]});
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    mt19937_64 rng(
        (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
    cin >> n;
    for (int i = 2; i <= n; i++) {
      if (!pr[i]) {
        ert[i] = rng();
        for (int j = i; j <= n; j += i) {
          pr[j] = i;
        }
      } else
        ert[i] = ert[i / pr[i]] ^ ert[pr[i]];
    }
    for (int i = 1; i <= n; i++) {
      ert[i] ^= ert[i - 1];
      xo ^= ert[i];
    }
    solve();
    xo ^= ert[n], n--;
    solve();
    return 0;
  }
  