#include <bits/stdc++.h>
using namespace std;
int f, t, m;
vector<int> ans;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> m;
  for (int i = 1; min(f, t) <= m; i++) {
    int j = i;
    while (j % 5 == 0) {
      j /= 5;
      f++;
    }
    j = i;
    while (j % 2 == 0) {
      j /= 2;
      t++;
    }
    if (min(f, t) == m) ans.push_back(i);
  }
  cout << ans.size() << endl;
  for (int i : ans) cout << i << " ";
  cout << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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 f, t, m;
vector<int> ans;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> m;
  for (int i = 1; min(f, t) <= m; i++) {
    int j = i;
    while (j % 5 == 0) {
      j /= 5;
      f++;
    }
    j = i;
    while (j % 2 == 0) {
      j /= 2;
      t++;
    }
    if (min(f, t) == m) ans.push_back(i);
  }
  cout << ans.size() << endl;
  for (int i : ans) cout << i << " ";
  cout << endl;
  return 0;
}

```