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 = 1000005;
  mt19937_64 rng(chrono::steady_clock().now().time_since_epoch().count());
  int n, lp[N];
  long long hsh[N];
  map<long long, int> mp;
  void out(vector<int> v) {
    for (auto x : v) lp[x] = -1;
    cout << n - v.size() << '\n';
    for (int i = 1; i <= n; ++i)
      if (lp[i] >= 0) cout << i << ' ';
    cout << '\n';
    exit(0);
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin >> n;
    for (int i = 2; i <= n; ++i) {
      if (lp[i]) continue;
      lp[i] = i;
      for (int j = i; j <= n; j += i)
        if (!lp[j]) lp[j] = i;
    }
    for (int i = 2; i <= n; ++i)
      if (lp[i] == i) hsh[i] = rng();
    for (int i = 2; i <= n; ++i) hsh[i] = hsh[i / lp[i]] ^ hsh[lp[i]];
    for (int i = 2; i <= n; ++i) hsh[i] ^= hsh[i - 1];
    long long tot = 0;
    for (int i = 1; i <= n; ++i) tot ^= hsh[i];
    if (tot == 0) out({});
    for (int i = 1; i <= n; ++i)
      if (tot == hsh[i]) out({i});
    for (int i = 1; i <= n; ++i) mp[hsh[i]] = i;
    for (int i = 1; i <= n; ++i)
      if (mp.count(tot ^ hsh[i])) out({i, mp[tot ^ hsh[i]]});
    tot ^= hsh[n];
    for (int i = 1; i <= n; ++i)
      if (mp.count(tot ^ hsh[i])) out({i, mp[tot ^ hsh[i]], n});
  }
  