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

### Prompt

Please formulate 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, x, i;
  cin >> n;
  for (i = 1; n > 0; i++) {
    x = i;
    while (x % 5 == 0) {
      x = x / 5;
      n = n - 1;
    }
  }
  if (n < 0) {
    cout << 0 << endl;
  }
  if (n == 0) {
    cout << 5 << endl;
    for (int j = i - 1; j <= i + 3; j++) {
      if (j == i - 1)
        cout << j;
      else
        cout << " " << j;
    }
  }
  cout << endl;
  return 0;
}

```