#include <bits/stdc++.h>
using namespace std;
int main() {
  int m;
  scanf("%d", &m);
  int num, x, sum = 0;
  for (int i = 1; i <= 100000; i++) {
    num = 1;
    x = i;
    while (1) {
      if (x % 5 == 0) {
        num++;
        x = x / 5;
      } else
        break;
    }
    sum += num;
    if (sum == m) {
      cout << "5"
           << "\n";
      for (int j = 5 * i; j < 5 * i + 5; j++) {
        printf("%d ", j);
      }
      return 0;
    } else if (sum > m) {
      break;
    }
  }
  cout << "0";
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
int main() {
  int m;
  scanf("%d", &m);
  int num, x, sum = 0;
  for (int i = 1; i <= 100000; i++) {
    num = 1;
    x = i;
    while (1) {
      if (x % 5 == 0) {
        num++;
        x = x / 5;
      } else
        break;
    }
    sum += num;
    if (sum == m) {
      cout << "5"
           << "\n";
      for (int j = 5 * i; j < 5 * i + 5; j++) {
        printf("%d ", j);
      }
      return 0;
    } else if (sum > m) {
      break;
    }
  }
  cout << "0";
  return 0;
}

```