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

```