#include <bits/stdc++.h>
using namespace std;
long long pot5(long long x) {
  int cont = 0;
  while (x % 5 == 0) {
    x /= 5;
    cont++;
  }
  return cont;
}
int main() {
  ios::sync_with_stdio(false);
  int m;
  cin >> m;
  {
    long long i;
    long long cont = 0;
    for (i = 5;; i += 5) {
      long long p = pot5(i);
      cont += p;
      if (cont >= m) break;
    }
    if (cont == m) {
      cout << 5 << endl;
      for (long long j = i; j < i + 5; ++j) {
        if (j != i) cout << " ";
        cout << j;
      }
      cout << endl;
    } else {
      cout << 0 << endl;
    }
  }
  return 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;
long long pot5(long long x) {
  int cont = 0;
  while (x % 5 == 0) {
    x /= 5;
    cont++;
  }
  return cont;
}
int main() {
  ios::sync_with_stdio(false);
  int m;
  cin >> m;
  {
    long long i;
    long long cont = 0;
    for (i = 5;; i += 5) {
      long long p = pot5(i);
      cont += p;
      if (cont >= m) break;
    }
    if (cont == m) {
      cout << 5 << endl;
      for (long long j = i; j < i + 5; ++j) {
        if (j != i) cout << " ";
        cout << j;
      }
      cout << endl;
    } else {
      cout << 0 << endl;
    }
  }
  return 0;
}

```