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;
  void FAST_IO(string filein = "", string fileout = "", string fileerr = "") {
    if (fopen(filein.c_str(), "r")) {
      freopen(filein.c_str(), "r", stdin);
      freopen(fileout.c_str(), "w", stdout);
    }
    cin.tie(0), cout.tie(0)->sync_with_stdio(0);
  }
  void Hollwo_Pelw();
  signed main() {
    FAST_IO("hollwo_pelw.inp", "hollwo_pelw.out");
    int testcases = 1;
    for (int test = 1; test <= testcases; test++) {
      Hollwo_Pelw();
    }
    return 0;
  }
  const int N = 1e6 + 5;
  mt19937_64 rng(
      (uint32_t)chrono::steady_clock::now().time_since_epoch().count());
  int n, p[N];
  uint64_t val[N], hsh[N], tot;
  inline void print(int n, vector<int> del) {
    cout << n - del.size() << '\n';
    for (int i = 1; i <= n; i++) {
      bool ok = 1;
      for (auto v : del)
        if (v == i) ok = 0;
      if (ok) cout << i << ' ';
    }
  }
  void Hollwo_Pelw() {
    cin >> n;
    iota(p + 1, p + n + 1, 1);
    for (int i = 2; i <= n; i++)
      if (p[i] == i)
        for (int j = 2 * i; j <= n; j += i) p[j] = i;
    for (int i = 2; i <= n; i++)
      if (p[i] == i) val[i] = rng();
    for (int i = 2; i <= n; i++) {
      hsh[i] = hsh[i / p[i]] ^ val[p[i]];
    }
    for (int i = 2; i <= n; i++) {
      hsh[i] ^= hsh[i - 1];
      tot ^= hsh[i];
    }
    if (tot == 0) return print(n, {});
    for (int i = 1; i <= n; i++)
      if (tot == hsh[i]) return print(n, {i});
    if (n % 2 == 0) return print(n, {2, n / 2});
    map<uint64_t, int> mp;
    for (int i = 1; i <= n; i++) {
      if (mp.count(hsh[i])) return print(n, {i, mp[hsh[i]]});
      mp[tot ^ hsh[i]] = i;
    }
    --n;
    print(n, {2, n / 2});
  }
  