#include <bits/stdc++.h>
using namespace std;
int m;
int up = 3000000, down = 0, mid = 200000;
int a[19], s;
int main() {
  cin >> m;
  a[0] = 1;
  for (int i = 1; i <= 10; i++) {
    a[i] = a[i - 1] * 5;
  }
  while (mid != down) {
    s = 0;
    for (int i = 1; i <= 10; i++) {
      s = s + (mid / a[i]);
    }
    if (s <= m) {
      down = mid;
      mid = (mid + up) / 2;
    } else {
      up = mid;
      mid = (mid + down) / 2;
    }
  }
  s = 0;
  for (int i = 1; i <= 10; i++) {
    s = s + (mid / a[i]);
  }
  if (s == m) {
    cout << 5 << endl;
    mid = (mid / 5) * 5;
    for (int i = 0; i < 5; i++) {
      cout << mid + i << ' ';
    }
  } else {
    cout << 0;
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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 m;
int up = 3000000, down = 0, mid = 200000;
int a[19], s;
int main() {
  cin >> m;
  a[0] = 1;
  for (int i = 1; i <= 10; i++) {
    a[i] = a[i - 1] * 5;
  }
  while (mid != down) {
    s = 0;
    for (int i = 1; i <= 10; i++) {
      s = s + (mid / a[i]);
    }
    if (s <= m) {
      down = mid;
      mid = (mid + up) / 2;
    } else {
      up = mid;
      mid = (mid + down) / 2;
    }
  }
  s = 0;
  for (int i = 1; i <= 10; i++) {
    s = s + (mid / a[i]);
  }
  if (s == m) {
    cout << 5 << endl;
    mid = (mid / 5) * 5;
    for (int i = 0; i < 5; i++) {
      cout << mid + i << ' ';
    }
  } else {
    cout << 0;
  }
  return 0;
}

```