#include <bits/stdc++.h>
using namespace std;
int main() {
  int temp = 0, n, coun = 0, chk = 1;
  cin >> n;
  for (int i = 5;; i++) {
    temp = 0;
    for (int j = 5; j <= i; j *= 5) {
      temp += i / j;
    }
    if (temp == n) {
      chk = i;
      break;
    } else if (temp > n) {
      chk = 0;
      break;
    }
  }
  if (chk)
    temp = chk, cout << 5 << endl
                     << temp << " " << temp + 1 << " " << temp + 2 << " "
                     << temp + 3 << " " << temp + 4 << 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;
int main() {
  int temp = 0, n, coun = 0, chk = 1;
  cin >> n;
  for (int i = 5;; i++) {
    temp = 0;
    for (int j = 5; j <= i; j *= 5) {
      temp += i / j;
    }
    if (temp == n) {
      chk = i;
      break;
    } else if (temp > n) {
      chk = 0;
      break;
    }
  }
  if (chk)
    temp = chk, cout << 5 << endl
                     << temp << " " << temp + 1 << " " << temp + 2 << " "
                     << temp + 3 << " " << temp + 4 << endl;
  else
    cout << 0 << endl;
  return 0;
}

```