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;
  int lpf[N];
  long long val[N];
  long long val1[N];
  long long val2[N];
  int use[N];
  int n;
  void finish() {
    cout << accumulate(use + 1, use + n + 1, 0) << '\n';
    for (int i = 1; i <= n; i++) {
      if (use[i]) cout << i << ' ';
    }
    cout << '\n';
    exit(0);
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    mt19937_64 rng(
        (long long)chrono::steady_clock::now().time_since_epoch().count());
    for (int i = 1; i < N; i++) {
      lpf[i] = i;
    }
    for (int i = 2; i < N; i++) {
      if (lpf[i] == i) {
        val[i] = rng();
        for (int j = 2 * i; j < N; j += i) {
          lpf[j] = i;
        }
      } else {
        val[i] = val[lpf[i]] ^ val[i / lpf[i]];
      }
    }
    for (int i = 1; i < N; i++) {
      val1[i] = val1[i - 1] ^ val[i];
      val2[i] = val2[i - 1] ^ val1[i];
    }
    cin >> n;
    for (int i = 1; i <= n; i++) use[i] = 1;
    long long x = val2[n];
    if (!x) finish();
    for (int i = 1; i <= n; i++) {
      if (val1[i] == x) {
        use[i] = 0;
        finish();
      }
    }
    unordered_map<long long, int> m;
    for (int i = 1; i <= n; i++) {
      if (m.count(val1[i] ^ x)) {
        use[i] = 0;
        use[m[val1[i] ^ x]] = 0;
        finish();
      }
      m[val1[i]] = i;
    }
    assert(n % 4 == 3);
    use[n] = 0;
    use[n / 2] = 0;
    use[2] = 0;
    finish();
  }
  