#include <bits/stdc++.h>
using namespace std;
void solve(int m) {
  int i = 0;
  int num = 0;
  while (num < m) {
    i += 5;
    int j = i;
    while (j % 5 == 0) {
      num++;
      j = j / 5;
    }
  }
  if (num != m) {
    cout << "0" << endl;
  } else {
    cout << "5" << endl;
    for (int j = i; j < i + 5; ++j) {
      cout << j << " ";
    }
    cout << endl;
  }
}
int main() {
  int m;
  cin >> m;
  solve(m);
  return 0;
}

### Prompt

Create a solution in cpp for 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;
void solve(int m) {
  int i = 0;
  int num = 0;
  while (num < m) {
    i += 5;
    int j = i;
    while (j % 5 == 0) {
      num++;
      j = j / 5;
    }
  }
  if (num != m) {
    cout << "0" << endl;
  } else {
    cout << "5" << endl;
    for (int j = i; j < i + 5; ++j) {
      cout << j << " ";
    }
    cout << endl;
  }
}
int main() {
  int m;
  cin >> m;
  solve(m);
  return 0;
}

```