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;
  bool check(long long x) {
    int y = int(sqrt(x) + 0.5);
    return 1ll * y * y == x;
  }
  vector<int> solve(int n) {
    vector<int> rt;
    if (n % 4 == 0) {
      rt.push_back(n / 2);
      return rt;
    }
    if (n % 4 == 1) {
      rt = solve(n - 1);
      rt.push_back(n);
      return rt;
    }
    if (n % 4 == 2) {
      if (check(n + 2)) {
        rt.push_back(n / 2 + 1);
        return rt;
      }
      rt.push_back(2);
      rt.push_back(n / 2);
      return rt;
    }
    rt = solve(n - 1);
    rt.push_back(n);
    if (check(1ll * n * (n - 3) / 2) && rt.size() > 2) {
      rt.clear();
      rt.push_back((n - 5) / 2);
      rt.push_back(n - 2);
    }
    return rt;
  }
  int main() {
    int n;
    scanf("%d", &n);
    if (n == 1) {
      printf("1\n1");
      return 0;
    }
    vector<int> de = solve(n);
    printf("%d\n", int(n - de.size()));
    for (int i = 1; i <= n; i++) {
      bool zd = false;
      for (int x : de) {
        if (i == x) {
          zd = true;
          break;
        }
      }
      if (!zd) printf("%d ", i);
    }
    return 0;
  }
  