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;
  mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
  mt19937 rnf(2106);
  const int N = 1000006, m = 60;
  int n;
  int u[N];
  long long h[N];
  long long p[N];
  bool solv(int n) {
    for (int i = 1; i <= n; ++i) u[i] = h[i] = p[i] = 0;
    u[1] = 1;
    for (int i = 2; i <= n; ++i) {
      if (u[i]) continue;
      for (int j = i; j <= n; j += i) {
        if (!u[j]) u[j] = i;
      }
    }
    for (int i = 1; i <= n; ++i) h[i] = rnd();
    for (int i = 1; i <= n; ++i) {
      int x = i;
      while (x > 1) {
        p[i] ^= h[u[x]];
        x /= u[x];
      }
    }
    for (int i = 1; i <= n; ++i) p[i] ^= p[i - 1];
    long long hh = 0;
    for (int i = 1; i <= n; ++i) hh ^= p[i];
    if (hh == 0) {
      cout << n << "\n";
      for (int i = 1; i <= n; ++i) cout << i << ' ';
      cout << "\n";
      return true;
    }
    for (int i = 1; i <= n; ++i) {
      if (p[i] == hh) {
        cout << n - 1 << "\n";
        for (int j = 1; j <= n; ++j) {
          if (i == j) continue;
          cout << j << ' ';
        }
        cout << "\n";
        return true;
      }
    }
    unordered_map<long long, int> mp;
    for (int i = 1; i <= n; ++i) mp[p[i]] = i;
    for (int i = 1; i <= n; ++i) {
      if (mp.find((hh ^ p[i])) != mp.end()) {
        cout << n - 2 << "\n";
        for (int j = 1; j <= n; ++j) {
          if (j == i || j == mp[(hh ^ p[i])]) continue;
          cout << j << ' ';
        }
        cout << "\n";
        return true;
      }
    }
    return false;
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int tt = 1;
    while (tt--) {
      int n;
      cin >> n;
      if (!solv(n)) solv(n - 1);
    }
    return 0;
  }
  