#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, ans, index, temp, k, num;
  ans = 0;
  cin >> n;
  num = n;
  index = 0;
  while (n > 0) {
    if (n <= 4) {
      ans += n * 5;
      n = 0;
      break;
    } else {
      ans += 4 * 5;
      n -= 4;
    }
    index++;
    if (index % 5 == 0) {
      ans += 5;
      temp = index;
      int count = 0;
      while (temp % 5 == 0) {
        temp /= 5;
        count++;
      }
      n -= (2 + count);
    } else {
      ans += 5;
      n -= 2;
    }
    if (n == 0 || n < 0) {
      break;
    }
  }
  if (n == 0) {
    cout << 5 << "\n";
    for (int i = 0; i <= 4; i++) {
      cout << ans + i << ' ';
    }
  } else {
    cout << 0;
  }
  return 0;
}

### Prompt

Please create a solution in CPP to 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() {
  long long n, ans, index, temp, k, num;
  ans = 0;
  cin >> n;
  num = n;
  index = 0;
  while (n > 0) {
    if (n <= 4) {
      ans += n * 5;
      n = 0;
      break;
    } else {
      ans += 4 * 5;
      n -= 4;
    }
    index++;
    if (index % 5 == 0) {
      ans += 5;
      temp = index;
      int count = 0;
      while (temp % 5 == 0) {
        temp /= 5;
        count++;
      }
      n -= (2 + count);
    } else {
      ans += 5;
      n -= 2;
    }
    if (n == 0 || n < 0) {
      break;
    }
  }
  if (n == 0) {
    cout << 5 << "\n";
    for (int i = 0; i <= 4; i++) {
      cout << ans + i << ' ';
    }
  } else {
    cout << 0;
  }
  return 0;
}

```