#include <bits/stdc++.h>
int a[1001000] = {0}, b[10];
int main(void) {
  int i, k = 0;
  for (i = 5; i < 1000010; i++) {
    if (i % 5 == 0) {
      int t = i;
      while (t % 5 == 0) {
        t /= 5;
        k++;
      }
    }
    a[i] = k;
  }
  int m, p = 0;
  scanf("%d", &m);
  for (i = 0; i < 1000010; i++) {
    if (a[i] == m) {
      b[p++] = i;
    }
  }
  printf("%d\n", p);
  for (i = 0; i < p; i++) {
    printf("%d ", b[i]);
  }
  return 0;
}

### Prompt

Please formulate a cpp solution to 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 a[1001000] = {0}, b[10];
int main(void) {
  int i, k = 0;
  for (i = 5; i < 1000010; i++) {
    if (i % 5 == 0) {
      int t = i;
      while (t % 5 == 0) {
        t /= 5;
        k++;
      }
    }
    a[i] = k;
  }
  int m, p = 0;
  scanf("%d", &m);
  for (i = 0; i < 1000010; i++) {
    if (a[i] == m) {
      b[p++] = i;
    }
  }
  printf("%d\n", p);
  for (i = 0; i < p; i++) {
    printf("%d ", b[i]);
  }
  return 0;
}

```