#include <bits/stdc++.h>
using namespace std;
int main() {
  int m;
  while (scanf("%d", &(m)) == 1) {
    vector<int> ans;
    for (int i = (5); i <= (500000); i++) {
      int sum = 0, t = i;
      while (t) {
        sum += t / 5;
        t /= 5;
      }
      if (sum == m) ans.push_back(i);
    }
    int sz = ans.size();
    printf("%d\n", sz);
    for (int i = (0); i <= (sz - 1); i++)
      printf("%d%c", ans[i], i == sz - 1 ? '\n' : ' ');
  }
  return 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() {
  int m;
  while (scanf("%d", &(m)) == 1) {
    vector<int> ans;
    for (int i = (5); i <= (500000); i++) {
      int sum = 0, t = i;
      while (t) {
        sum += t / 5;
        t /= 5;
      }
      if (sum == m) ans.push_back(i);
    }
    int sz = ans.size();
    printf("%d\n", sz);
    for (int i = (0); i <= (sz - 1); i++)
      printf("%d%c", ans[i], i == sz - 1 ? '\n' : ' ');
  }
  return 0;
}

```