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 n, a[400015], b[400015], c[400015], ans[50], cnt;
int main() {
  for (int i = 1; i <= 400010; i++) {
    int x = 0, y = 0, t = i;
    while (t % 2 == 0) t /= 2, x++;
    while (t % 5 == 0) t /= 5, y++;
    a[i] = a[i - 1] + x, b[i] = b[i - 1] + y, c[i] = min(a[i], b[i]);
  }
  cin >> n;
  for (int i = 1; i <= 400010; i++) {
    if (c[i] == n) {
      ans[++cnt] = i;
    }
    if (c[i] > n) break;
  }
  cout << cnt << endl;
  for (int i = 1; i <= cnt; i++) cout << ans[i] << " ";
  return 0;
}