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;
  template <class T>
  inline bool mnto(T& a, T b) {
    return a > b ? a = b, 1 : 0;
  }
  template <class T>
  inline bool mxto(T& a, T b) {
    return a < b ? a = b, 1 : 0;
  }
  int n;
  int lp[1000005];
  long long hsh[1000005];
  vector<int> primes;
  map<long long, int> mp;
  int main() {
    mt19937_64 rnd(234901);
    scanf("%d", &n);
    for (int i = 2; i < n + 1; i++) {
      if (lp[i] == 0) {
        lp[i] = i;
        hsh[i] = rnd();
        primes.push_back(i);
      };
      for (int j = 0; j < primes.size() && primes[j] <= lp[i] &&
                      (long long)primes[j] * i <= n;
           j++) {
        int x = primes[j] * i;
        lp[x] = primes[j];
        hsh[x] = hsh[primes[j]] ^ hsh[i];
        ;
      }
    }
    long long tot = 0;
    for (int i = 2; i < n + 1; i++) {
      hsh[i] ^= hsh[i - 1];
      tot ^= hsh[i];
      mp[hsh[i]] = i;
      ;
    };
    if (tot == 0) {
      printf("%d\n", n);
      for (int i = 1; i < n + 1; i++) {
        printf("%d ", i);
      }
      printf("\n");
      return 0;
    }
    for (int i = 1; i < n + 1; i++) {
      if ((tot ^ hsh[i]) == 0) {
        printf("%d\n", n - 1);
        for (int j = 1; j < n + 1; j++) {
          if (j == i) continue;
          printf("%d ", j);
        }
        printf("\n");
        return 0;
      }
    }
    for (int i = 1; i < n + 1; i++) {
      if (mp[tot ^ hsh[i]]) {
        assert(tot ^ hsh[i] ^ hsh[mp[tot ^ hsh[i]]] == 0);
        printf("%d\n", n - 2);
        for (int j = 1; j < n + 1; j++) {
          if (j == i || j == mp[tot ^ hsh[i]]) continue;
          printf("%d ", j);
        }
        printf("\n");
        return 0;
      }
    }
    printf("%d\n", n - 3);
    for (int i = 1; i < n + 1; i++) {
      if (i == 2 || i == (n - 1) / 2 || i == n) continue;
      printf("%d ", i);
    }
    printf("\n");
    return 0;
  }
  