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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const double PI = acos(-1.0);
const double EPS = 1e-12;
inline int read() {
  static char ch;
  bool sgn = false;
  while (ch = getchar(), ch < '0' || ch > '9')
    if (ch == '-') sgn = true;
  int res = ch - 48;
  while (ch = getchar(), ch >= '0' && ch <= '9') res = res * 10 + ch - 48;
  return sgn ? -res : res;
}
int ans[1000000];
int main() {
  int m = read(), x = 0, y = 0;
  for (int i = 1; i <= 1000000; i++) {
    if (i % 2 == 0) {
      int t = i;
      while (t % 2 == 0) {
        x++;
        t /= 2;
      }
    }
    if (i % 5 == 0) {
      int t = i;
      while (t % 5 == 0) {
        y++;
        t /= 5;
      }
    }
    if ((x < y ? x : y) == m) {
      ans[0]++;
      ans[ans[0]] = i;
    }
    if ((x < y ? x : y) > m) break;
  }
  printf("%d\n", ans[0]);
  for (int i = 1; i <= ans[0]; i++) printf("%d ", ans[i]);
}