#include <bits/stdc++.h>
using namespace std;
int calculate(int n) {
  int sum = 0;
  while ((n / 5) != 0) {
    sum += n / 5;
    n /= 5;
  }
  return sum;
}
int main() {
  int k, i, s, var;
  cin >> k;
  i = 0;
  while (1) {
    s = calculate(5 * i);
    if (k == s) break;
    if (k < s) {
      i = 0;
      break;
    }
    i++;
  }
  if (i != 0) {
    printf("%d\n", 5);
    for (var = 0; var < 5; var++) printf("%d ", 5 * i + var);
  } else
    printf("0");
  printf("\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>
using namespace std;
int calculate(int n) {
  int sum = 0;
  while ((n / 5) != 0) {
    sum += n / 5;
    n /= 5;
  }
  return sum;
}
int main() {
  int k, i, s, var;
  cin >> k;
  i = 0;
  while (1) {
    s = calculate(5 * i);
    if (k == s) break;
    if (k < s) {
      i = 0;
      break;
    }
    i++;
  }
  if (i != 0) {
    printf("%d\n", 5);
    for (var = 0; var < 5; var++) printf("%d ", 5 * i + var);
  } else
    printf("0");
  printf("\n");
  return 0;
}

```