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>
int main() {
  int n, k = 5, i, x[100] = {}, gs = 0, y[100];
  scanf("%d", &n);
  for (i = 1; i <= 100000; i = i * 5 + 1, k *= 5) {
    if (i > n) break;
    x[gs] = i;
    y[gs++] = k;
  }
  int ans = 0, remain = n, j;
  for (i = gs - 1; i >= 0; i--) {
    for (j = 0; j <= 4; j++)
      if (x[i] * j <= remain && x[i] * (j + 1) > remain) break;
    if (j > 4) break;
    ans += y[i] * j;
    remain -= x[i] * j;
  }
  if (remain != 0)
    printf("0\n");
  else
    printf("5\n%d %d %d %d %d\n", ans, ans + 1, ans + 2, ans + 3, ans + 4);
  return 0;
}