#include <bits/stdc++.h>
using namespace std;
int n;
int x, y;
int main() {
  while (scanf("%d", &n) != EOF) {
    int ans = 5, cnt = 1, tmp;
    while (cnt < n) {
      ans += 5;
      tmp = ans;
      while (tmp % 5 == 0) {
        tmp /= 5;
        cnt++;
      }
    }
    if (cnt == n) {
      printf("5\n");
      for (int i = ans; i < ans + 5; i++) {
        printf("%d", i);
        if (i < ans + 4)
          printf(" ");
        else
          puts("");
      }
    } else
      puts("0");
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n;
int x, y;
int main() {
  while (scanf("%d", &n) != EOF) {
    int ans = 5, cnt = 1, tmp;
    while (cnt < n) {
      ans += 5;
      tmp = ans;
      while (tmp % 5 == 0) {
        tmp /= 5;
        cnt++;
      }
    }
    if (cnt == n) {
      printf("5\n");
      for (int i = ans; i < ans + 5; i++) {
        printf("%d", i);
        if (i < ans + 4)
          printf(" ");
        else
          puts("");
      }
    } else
      puts("0");
  }
  return 0;
}

```