Problem: Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmers who can solve this problem?

Input

The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.

Output

First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.

Examples

Input

1


Output

5
5 6 7 8 9 

Input

5


Output

0

Note

The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·...·n.

In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int p(int n) {
  int s = 0, i = 5;
  while (n >= i) {
    s += n / i;
    i *= 5;
  }
  return s;
}
int main() {
  int m;
  int a[200000];
  a[1] = 1;
  cin >> m;
  if (m == 1) {
    cout << 5 << "\n";
    for (int i = 5; i <= 9; i++) {
      cout << i << " ";
    }
    return 0;
  }
  for (int i = 2; i <= 1e10; i++) {
    a[i] = p(5 * i);
    if (a[i] > m) {
      cout << 0;
      return 0;
    }
    if (a[i] == m) {
      cout << 5 << "\n";
      for (int j = 5 * i; j <= 5 * i + 4; j++) {
        cout << j << " ";
      }
      return 0;
    }
  }
}