#include <bits/stdc++.h>
using namespace std;
int a(int n) {
  int i;
  while (n % 5 == 0) {
    i++;
    n /= 5;
  }
  return i;
}
int main(int argc, char** argv) {
  int n = 0, k = 5, i = 0, m;
  scanf("%d", &m);
  while (1) {
    i += a(k);
    if (i == m) {
      n = 1;
      break;
    }
    if (i > m) break;
    k += 5;
  }
  if (n == 1) {
    printf("5\n");
    printf("%d %d %d %d %d", k, k + 1, k + 2, k + 3, k + 4);
  }
  if (n == 0) printf("0");
  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>
using namespace std;
int a(int n) {
  int i;
  while (n % 5 == 0) {
    i++;
    n /= 5;
  }
  return i;
}
int main(int argc, char** argv) {
  int n = 0, k = 5, i = 0, m;
  scanf("%d", &m);
  while (1) {
    i += a(k);
    if (i == m) {
      n = 1;
      break;
    }
    if (i > m) break;
    k += 5;
  }
  if (n == 1) {
    printf("5\n");
    printf("%d %d %d %d %d", k, k + 1, k + 2, k + 3, k + 4);
  }
  if (n == 0) printf("0");
  return 0;
}

```