#include <bits/stdc++.h>
int x, i, flag, flag2, a, sayac, b;
int main() {
  scanf("%d", &x);
  a = 5;
  while (1) {
    b = a;
    while (b) {
      if (b % 5 != 0)
        break;
      else {
        sayac++;
        b = b / 5;
      }
    }
    if (sayac == x) {
      printf("5\n");
      for (int i = a; i < a + 5; i++) {
        printf("%d ", i);
      }
      return 0;
    }
    if (sayac > x) {
      printf("0");
      return 0;
    }
    a = a + 5;
  }
  return 0;
}

### Prompt

Create a solution in cpp for the following 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.

### Response

```cpp
#include <bits/stdc++.h>
int x, i, flag, flag2, a, sayac, b;
int main() {
  scanf("%d", &x);
  a = 5;
  while (1) {
    b = a;
    while (b) {
      if (b % 5 != 0)
        break;
      else {
        sayac++;
        b = b / 5;
      }
    }
    if (sayac == x) {
      printf("5\n");
      for (int i = a; i < a + 5; i++) {
        printf("%d ", i);
      }
      return 0;
    }
    if (sayac > x) {
      printf("0");
      return 0;
    }
    a = a + 5;
  }
  return 0;
}

```