#include <bits/stdc++.h>
int main() {
  int a, b, c, n, i = 0, j = 0, t;
  scanf("%d", &n);
  while (5 * n - i * 5 > 0) {
    c = 0;
    t = 5 * (n - i);
    while (t) {
      c += t / 5;
      t = t / 5;
    }
    if (c == n) {
      printf("5\n");
      for (j = 0; j < 5; j++) printf("%d ", 5 * (n - i) + j);
      return 0;
    }
    if (c < n) {
      printf("0");
      return 0;
    }
    i++;
  }
}

### 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>
int main() {
  int a, b, c, n, i = 0, j = 0, t;
  scanf("%d", &n);
  while (5 * n - i * 5 > 0) {
    c = 0;
    t = 5 * (n - i);
    while (t) {
      c += t / 5;
      t = t / 5;
    }
    if (c == n) {
      printf("5\n");
      for (j = 0; j < 5; j++) printf("%d ", 5 * (n - i) + j);
      return 0;
    }
    if (c < n) {
      printf("0");
      return 0;
    }
    i++;
  }
}

```