#include <bits/stdc++.h>
using namespace std;
const int maxn = 10;
int f[maxn + 1][maxn + 1];
int main(int argc, char *argv[]) {
  int m;
  cin >> m;
  int i = 0, ans = 0;
  while (m > 0) {
    i += 5;
    int j = i;
    while (j % 5 == 0) {
      j /= 5;
      m -= 1;
    }
  }
  if (m == 0) {
    cout << 5 << endl
         << i << ' ' << (i + 1) << ' ' << (i + 2) << ' ' << (i + 3) << ' '
         << (i + 4) << endl;
  } else
    cout << 0 << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const int maxn = 10;
int f[maxn + 1][maxn + 1];
int main(int argc, char *argv[]) {
  int m;
  cin >> m;
  int i = 0, ans = 0;
  while (m > 0) {
    i += 5;
    int j = i;
    while (j % 5 == 0) {
      j /= 5;
      m -= 1;
    }
  }
  if (m == 0) {
    cout << 5 << endl
         << i << ' ' << (i + 1) << ' ' << (i + 2) << ' ' << (i + 3) << ' '
         << (i + 4) << endl;
  } else
    cout << 0 << endl;
  return 0;
}

```