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>
const int N = 123456;
int v[N];
inline int go(int x) {
  int ans = 0;
  while (0 < x) {
    x /= 5;
    ans += x;
  }
  return ans;
}
int main() {
  int m;
  scanf("%d", &m);
  int l = 0, r = int(1e9);
  while (l <= r) {
    int mid = (l + r) / 2;
    if (go(mid) < m) {
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  int ans = 0;
  while (go(l) == m) {
    v[ans] = l;
    ++ans;
    ++l;
  }
  printf("%d\n", ans);
  for (int i = 0; i < ans; ++i) {
    if (0 < i) {
      putchar(' ');
    }
    printf("%d", v[i]);
  }
  if (0 < ans) {
    putchar('\n');
  }
  return 0;
}