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 ll = long long;
  using namespace std;
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    int n;
    cin >> n;
    vector<int> prime(n + 1, 0);
    for (int i = 2; i <= n; i++) {
      if (!prime[i]) {
        for (int j = i; j <= n; j += i) prime[j] = i;
      }
    }
    mt19937_64 rng(233);
    vector<long long> h(n + 1);
    for (int i = 2; i <= n; i++)
      if (prime[i] == i) h[i] = rng();
    long long P = 0;
    vector<long long> p(n + 1);
    for (int i = 1; i <= n; i++) {
      p[i] = p[i - 1];
      int x = i;
      while (x > 1) {
        p[i] ^= h[prime[x]];
        x /= prime[x];
      }
      P ^= p[i];
    }
    auto print = [&](vector<int> erased) {
      cout << n - erased.size() << '\n';
      for (int i = 1; i <= n; i++)
        if (!count(erased.begin(), erased.end(), i)) cout << i << ' ';
      cout << '\n';
      exit(0);
    };
    if (P == 0) print({});
    for (int i = 1; i <= n; i++)
      if (p[i] == P) print({i});
    map<long long, int> mp;
    for (int i = 1; i <= n; i++) {
      if (mp.count(P ^ p[i])) print({i, mp[P ^ p[i]]});
      mp[p[i]] = i;
    }
    assert(n % 2 == 1);
    print({2, (n - 1) / 2, n});
    return 0;
  }
  