#include <bits/stdc++.h>
using namespace std;
void solve();
int main() {
  int m = 1;
  for (int i = 0; i < m; i++) {
    solve();
  }
  return 0;
}
void solve() {
  long long m;
  cin >> m;
  long long n = 5;
  long long k = 1;
  vector<long long> ans;
  while (k <= m) {
    if (k == m) {
      ans.push_back(n);
    }
    long long r = n + 1;
    while (r % 5 == 0) {
      k++;
      r /= 5;
    }
    n++;
  }
  cout << ans.size() << endl;
  for (long long i = 0; i < ans.size(); i++) {
    cout << ans[i] << ' ';
  }
}

### 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;
void solve();
int main() {
  int m = 1;
  for (int i = 0; i < m; i++) {
    solve();
  }
  return 0;
}
void solve() {
  long long m;
  cin >> m;
  long long n = 5;
  long long k = 1;
  vector<long long> ans;
  while (k <= m) {
    if (k == m) {
      ans.push_back(n);
    }
    long long r = n + 1;
    while (r % 5 == 0) {
      k++;
      r /= 5;
    }
    n++;
  }
  cout << ans.size() << endl;
  for (long long i = 0; i < ans.size(); i++) {
    cout << ans[i] << ' ';
  }
}

```