#include <bits/stdc++.h>
using namespace std;
long long m, k, h, n, i;
int main() {
  cin >> m;
  for (i = 5;; i += 5) {
    h = i;
    while (h % 5 == 0) {
      n++;
      h /= 5;
    }
    if (n > m) {
      cout << 0;
      return 0;
    } else if (n == m) {
      break;
    }
  }
  cout << 5 << endl;
  cout << i << " " << i + 1 << " " << i + 2 << " " << i + 3 << " " << i + 4;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long long m, k, h, n, i;
int main() {
  cin >> m;
  for (i = 5;; i += 5) {
    h = i;
    while (h % 5 == 0) {
      n++;
      h /= 5;
    }
    if (n > m) {
      cout << 0;
      return 0;
    } else if (n == m) {
      break;
    }
  }
  cout << 5 << endl;
  cout << i << " " << i + 1 << " " << i + 2 << " " << i + 3 << " " << i + 4;
  return 0;
}

```