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 main() {
  int m;
  cin >> m;
  int cnt = 0;
  int i = 0;
  for (i = 1; cnt < m; ++i) {
    int temp = i;
    cnt += 1;
    while (temp % 5 == 0) {
      cnt += 1;
      temp /= 5;
    }
  }
  i -= 1;
  if (cnt != m)
    cout << 0 << endl;
  else {
    cout << 5 << endl;
    printf("%d %d %d %d %d\n", 5 * i, 5 * i + 1, 5 * i + 2, 5 * i + 3,
           5 * i + 4);
  }
}