#include <bits/stdc++.h>
int max(int a, int b) { return a < b ? a : b; }
int num2(int m) { return m % 2 ? 0 : 1 + num2(m / 2); }
int num5(int m) { return m % 5 ? 0 : 1 + num5(m / 5); }
int main() {
  int m, i;
  while (~scanf("%d", &m)) {
    int a[2] = {0, 0};
    int c, b;
    c = b = 0;
    int j = 0;
    for (i = 1;; i++) {
      c += num2(i);
      b += num5(i);
      if (max(c, b) == m) {
        if (!j)
          a[j] = a[j + 1] = i, j = 1;
        else
          a[j] = i;
      }
      if (max(c, b) > m) break;
    }
    if (a[0]) {
      printf("%d\n", a[1] - a[0] + 1);
      for (; a[0] < a[1]; a[0]++) printf("%d ", a[0]);
      printf("%d\n", a[1]);
    } else
      printf("0\n");
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
int max(int a, int b) { return a < b ? a : b; }
int num2(int m) { return m % 2 ? 0 : 1 + num2(m / 2); }
int num5(int m) { return m % 5 ? 0 : 1 + num5(m / 5); }
int main() {
  int m, i;
  while (~scanf("%d", &m)) {
    int a[2] = {0, 0};
    int c, b;
    c = b = 0;
    int j = 0;
    for (i = 1;; i++) {
      c += num2(i);
      b += num5(i);
      if (max(c, b) == m) {
        if (!j)
          a[j] = a[j + 1] = i, j = 1;
        else
          a[j] = i;
      }
      if (max(c, b) > m) break;
    }
    if (a[0]) {
      printf("%d\n", a[1] - a[0] + 1);
      for (; a[0] < a[1]; a[0]++) printf("%d ", a[0]);
      printf("%d\n", a[1]);
    } else
      printf("0\n");
  }
  return 0;
}

```