#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i = 5, temp = 0;
  int count = 0;
  int n;
  cin >> n;
  while (count < n) {
    count = 0;
    i = 5;
    while (i <= temp) {
      count += temp / i;
      i *= 5;
    }
    if (count < n) {
      temp += 5;
    }
  }
  if (n == count) {
    cout << 5 << '\n';
    for (int j = 0; j < 5; j++) {
      cout << temp + j << ' ';
    }
  } else {
    cout << 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i = 5, temp = 0;
  int count = 0;
  int n;
  cin >> n;
  while (count < n) {
    count = 0;
    i = 5;
    while (i <= temp) {
      count += temp / i;
      i *= 5;
    }
    if (count < n) {
      temp += 5;
    }
  }
  if (n == count) {
    cout << 5 << '\n';
    for (int j = 0; j < 5; j++) {
      cout << temp + j << ' ';
    }
  } else {
    cout << 0;
  }
}

```