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 tzero(int n) {
  int count = 0;
  for (int i = 5; n / i >= 1; i *= 5) count += n / i;
  return count;
}
int main() {
  long long int m, n, i, j, k, ans = 0, a[500006], t = 0, hop;
  cin >> m;
  for (i = 0; i <= 500000; i++) {
    a[i] = i + 1;
  }
  for (i = 500000; i >= 1; i--) {
    if (tzero(i) == m) {
      hop = i;
      break;
    }
  }
  for (i = 1; i <= hop; i++) {
    if (tzero(i) == m) {
      ans++;
      a[t++] = i;
    }
  }
  cout << ans << " ";
  for (i = 0; i < t; i++) {
    cout << a[i] << " ";
  }
}