#include <bits/stdc++.h>
using namespace std;
long long p5(long long a) {
  long long num = 5, ret = 0;
  while (a / num > 0) {
    ret += a / num;
    num *= 5;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int m;
  cin >> m;
  long long l = 0, r = 5;
  while (p5(l) < m) {
    l += 5;
    r += 5;
  }
  if (p5(l) > m)
    cout << 0;
  else {
    cout << r - l << "\n";
    for (int i = l; i < r; ++i) {
      cout << i << " ";
    }
  }
}

### 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;
long long p5(long long a) {
  long long num = 5, ret = 0;
  while (a / num > 0) {
    ret += a / num;
    num *= 5;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int m;
  cin >> m;
  long long l = 0, r = 5;
  while (p5(l) < m) {
    l += 5;
    r += 5;
  }
  if (p5(l) > m)
    cout << 0;
  else {
    cout << r - l << "\n";
    for (int i = l; i < r; ++i) {
      cout << i << " ";
    }
  }
}

```