#include <bits/stdc++.h>
using namespace std;
long long m;
long long temp;
int flag = 0;
int i;
const long long N = 40000000;
int main() {
  ios::sync_with_stdio(0);
  cin >> m;
  for (i = 0; i <= N; i += 5) {
    temp = i / 390625 + i / 78125 + i / 15625 + i / 3125 + i / 625 + i / 125 +
           i / 25 + i / 5;
    if (temp == m) {
      flag = 1;
      break;
    }
    if (temp > m) {
      break;
    }
  }
  if (flag) {
    cout << 5 << endl;
    cout << i << ' ' << i + 1 << ' ' << i + 2 << ' ' << i + 3 << ' ' << i + 4
         << endl;
  } else
    cout << 0 << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 m;
long long temp;
int flag = 0;
int i;
const long long N = 40000000;
int main() {
  ios::sync_with_stdio(0);
  cin >> m;
  for (i = 0; i <= N; i += 5) {
    temp = i / 390625 + i / 78125 + i / 15625 + i / 3125 + i / 625 + i / 125 +
           i / 25 + i / 5;
    if (temp == m) {
      flag = 1;
      break;
    }
    if (temp > m) {
      break;
    }
  }
  if (flag) {
    cout << 5 << endl;
    cout << i << ' ' << i + 1 << ' ' << i + 2 << ' ' << i + 3 << ' ' << i + 4
         << endl;
  } else
    cout << 0 << endl;
  return 0;
}

```