#include <bits/stdc++.h>
using namespace std;
vector<int> ans;
int m, n = 1e6;
int main() {
  cin >> m;
  while (n--) {
    int s = 0, times = 10, pw = 5;
    while (times--) {
      s += n / pw;
      pw *= 5;
    }
    if (m == s) ans.push_back(n);
  }
  printf("%d\n", ans.size());
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  return 0;
}

### Prompt

Generate 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;
vector<int> ans;
int m, n = 1e6;
int main() {
  cin >> m;
  while (n--) {
    int s = 0, times = 10, pw = 5;
    while (times--) {
      s += n / pw;
      pw *= 5;
    }
    if (m == s) ans.push_back(n);
  }
  printf("%d\n", ans.size());
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  return 0;
}

```