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 INF = (int)(1e9) + 5, MN = (int)(1e6) + 5;
  int n, tab[MN];
  unsigned long long h[MN], f[MN];
  map<unsigned long long, int> dict;
  void solve() {
    cin >> n;
    for (int i = 2; i <= n; i++) {
      if (!tab[i]) {
        for (int j = 2 * i; j <= n; j += i) {
          if (!tab[j]) {
            tab[j] = i;
          }
        }
      }
    }
    mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
    for (int i = 2; i <= n; i++) {
      if (!tab[i]) {
        h[i] = rnd();
      } else {
        h[i] = h[tab[i]] ^ h[i / tab[i]];
      }
    }
    unsigned long long sum_h = 0;
    dict[0] = 1;
    for (int i = 2; i <= n; i++) {
      f[i] = f[i - 1] ^ h[i];
      dict[f[i]] = i;
      sum_h ^= f[i];
    }
    if (sum_h == 0) {
      cout << n << "\n";
      for (int i = 1; i <= n; i++) {
        cout << i << " ";
      }
      return;
    }
    if (dict.find(sum_h) != dict.end()) {
      cout << n - 1 << "\n";
      int del = dict.find(sum_h)->second;
      for (int i = 1; i <= n; i++) {
        if (i != del) {
          cout << i << " ";
        }
      }
      return;
    }
    for (int i = 1; i <= n; i++) {
      if (dict.find(sum_h ^ f[i]) != dict.end() &&
          dict.find(sum_h ^ f[i])->second != i) {
        cout << n - 2 << "\n";
        int del = dict.find(sum_h ^ f[i])->second;
        for (int j = 1; j <= n; j++) {
          if (j != i && j != del) {
            cout << j << " ";
          }
        }
        return;
      }
    }
    cout << n - 3 << "\n";
    for (int i = 1; i <= n; i++) {
      if (i != 2 && i != n && i != (n - 1) / 2) {
        cout << i << " ";
      }
    }
  }
  void clear() { cout << "\n"; }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int test = 1;
    while (test--) {
      solve();
      clear();
    }
    return 0;
  }
  