#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  long long a[15];
  cin >> n;
  int n1 = 0;
  a[0] = 1;
  for (int k = 1; k <= 14; k++) {
    a[k] = a[k - 1] * 5;
  }
  long long koe = 0;
  while (n1 < n) {
    koe++;
    for (int k = 14; k >= 0; k--) {
      if (koe % a[k] == 0) {
        n1 += k + 1;
        break;
      }
    }
  }
  if (n1 == n) {
    cout << "5" << endl;
    for (int k = koe * 5; k < koe * 5 + 5; k++) cout << k << " ";
  } else
    cout << "0";
  return 0;
}

### Prompt

Your challenge is to write 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 n;
  long long a[15];
  cin >> n;
  int n1 = 0;
  a[0] = 1;
  for (int k = 1; k <= 14; k++) {
    a[k] = a[k - 1] * 5;
  }
  long long koe = 0;
  while (n1 < n) {
    koe++;
    for (int k = 14; k >= 0; k--) {
      if (koe % a[k] == 0) {
        n1 += k + 1;
        break;
      }
    }
  }
  if (n1 == n) {
    cout << "5" << endl;
    for (int k = koe * 5; k < koe * 5 + 5; k++) cout << k << " ";
  } else
    cout << "0";
  return 0;
}

```