#include <bits/stdc++.h>
using namespace std;
int main() {
  int m;
  int n, x;
  bool encontrado = false;
  cin >> m;
  for (int i = 0; i < 500000; i++) {
    x = (i / 5 + i / 25 + i / 125 + i / 625 + i / 3125 + i / 15625 + i / 78125 +
         i / 390625);
    if (x == m) {
      cout << "5" << endl;
      n = i;
      encontrado = true;
      break;
    }
  }
  if (encontrado == true) {
    cout << n << " " << n + 1 << " " << n + 2 << " " << n + 3 << " " << n + 4;
  } else {
    cout << "0";
  }
}

### Prompt

In Cpp, your task is to solve 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;
  int n, x;
  bool encontrado = false;
  cin >> m;
  for (int i = 0; i < 500000; i++) {
    x = (i / 5 + i / 25 + i / 125 + i / 625 + i / 3125 + i / 15625 + i / 78125 +
         i / 390625);
    if (x == m) {
      cout << "5" << endl;
      n = i;
      encontrado = true;
      break;
    }
  }
  if (encontrado == true) {
    cout << n << " " << n + 1 << " " << n + 2 << " " << n + 3 << " " << n + 4;
  } else {
    cout << "0";
  }
}

```