#include <bits/stdc++.h>
using namespace std;
int check(int n) {
  int ct = 0;
  while (n) {
    n /= 5;
    ct += n;
  }
  return ct;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int m;
  cin >> m;
  vector<int> a;
  int i = 1, cnt = 0;
  while (i > 0) {
    if (check(i) == m) {
      cnt++;
      a.push_back(i);
    }
    if (check(i) > m) {
      break;
    }
    i++;
  }
  cout << cnt << "\n";
  ;
  if (a.size() > 0)
    for (auto it = a.begin(); it != a.end(); it++) cout << *it << ' ';
  cout << '\n';
  ;
}

### Prompt

Your challenge is to write 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;
int check(int n) {
  int ct = 0;
  while (n) {
    n /= 5;
    ct += n;
  }
  return ct;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int m;
  cin >> m;
  vector<int> a;
  int i = 1, cnt = 0;
  while (i > 0) {
    if (check(i) == m) {
      cnt++;
      a.push_back(i);
    }
    if (check(i) > m) {
      break;
    }
    i++;
  }
  cout << cnt << "\n";
  ;
  if (a.size() > 0)
    for (auto it = a.begin(); it != a.end(); it++) cout << *it << ' ';
  cout << '\n';
  ;
}

```