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;
  int main() {
    int n;
    scanf("%d", &n);
    if (n == 1) {
      printf("1\n1\n");
      return 0;
    }
    if (n & 1) {
      long long prod = 1ll * n * ((n >> 1) - 1);
      long long s = sqrt(prod);
      if (s * s == prod) {
        printf("%d\n", n - 2);
        int s1 = (n >> 1) - 2, s2 = n - 2;
        for (int i = 1; i <= n; i++)
          if (i != s1 && i != s2) printf("%d ", i);
        putchar('\n');
        return 0;
      }
    }
    if (n & 1) n--;
    int m = n >> 1;
    if (m & 1) {
      int x = 0, k = (m + 1) >> 1, s = sqrt(k);
      if (s * s == k)
        x = m + 1;
      else if (m == 9)
        x = 7;
      if (x) {
        printf("%d\n", n - 1);
        for (int i = 1; i <= n; i++)
          if (i != x) printf("%d ", i);
        putchar('\n');
      } else {
        printf("%d\n", n - 2);
        for (int i = 1; i <= n; i++)
          if (i != 2 && i != m) printf("%d ", i);
        putchar('\n');
      }
    } else {
      printf("%d\n", n - 1);
      for (int i = 1; i <= n; i++)
        if (i != m) printf("%d ", i);
      putchar('\n');
    }
    return 0;
  }
  