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;
  constexpr long long MOD = 998244353;
  int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    long long n;
    cin >> n;
    mt19937_64 rng((int)chrono::steady_clock::now().time_since_epoch().count());
    map<long long, long long> prime_hash;
    vector<long long> n_hash(n + 1), f_hash(n + 1);
    vector<int> min_primes(n + 1), primes;
    primes.reserve(1e5);
    for (int i = 2; i <= n; i++) {
      if (!min_primes[i])
        min_primes[i] = i, primes.push_back(i), prime_hash[i] = rng();
      for (auto& p : primes) {
        if (i * p > n) break;
        min_primes[i * p] = p;
        if (i % p == 0) break;
      }
    }
    long long rolling = 0;
    for (int i = 2; i <= n; i++) {
      int x = i;
      while (x > 1) {
        n_hash[i] ^= prime_hash[min_primes[x]];
        x /= min_primes[x];
      }
      f_hash[i] = n_hash[i] ^ f_hash[i - 1];
      rolling ^= f_hash[i];
    }
    map<long long, long long> inv_f_hash;
    for (int i = 2; i <= n; i++) inv_f_hash[f_hash[i]] = i;
    auto get_wont = [&]() -> vector<long long> {
      if (rolling == 0) {
        return {};
      } else if (inv_f_hash.count(rolling)) {
        return {inv_f_hash[rolling]};
      } else {
        for (int k = 2; k <= n; k++)
          if (inv_f_hash.count(rolling ^ f_hash[k]))
            return {k, inv_f_hash[rolling ^ f_hash[k]]};
        return {2, n / 2, n};
      }
    };
    auto wont = get_wont();
    cout << n - wont.size() << "\n";
    for (int i = 1; i <= n; i++)
      if (find(wont.begin(), wont.end(), i) == wont.end()) cout << i << " ";
    cout << "\n";
  }
  