#include <bits/stdc++.h>
using namespace std;
long long m, i, ans, k, j, a[1000002], b[1000002];
int main() {
  cin >> m;
  for (i = 1; i <= 1000000; i++) {
    k = 1;
    for (j = 1; j <= 10; j++) {
      k *= 5;
      a[i] += i / k;
    }
    if (a[i] == m) {
      ans++;
      b[ans] = i;
    }
    if (ans == 5) break;
  }
  cout << ans << endl;
  for (i = 1; i <= ans; i++) cout << b[i] << " ";
}

### Prompt

Please formulate 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;
long long m, i, ans, k, j, a[1000002], b[1000002];
int main() {
  cin >> m;
  for (i = 1; i <= 1000000; i++) {
    k = 1;
    for (j = 1; j <= 10; j++) {
      k *= 5;
      a[i] += i / k;
    }
    if (a[i] == m) {
      ans++;
      b[ans] = i;
    }
    if (ans == 5) break;
  }
  cout << ans << endl;
  for (i = 1; i <= ans; i++) cout << b[i] << " ";
}

```