#include <bits/stdc++.h>
using namespace std;
int main() {
  long long m;
  cin >> m;
  long long n = 1000000, count = 0, ans = 0;
  for (long long j = 5; j < n; j++) {
    count = 0;
    for (long long i = 5; j / i > 0; i = i * 5) count += j / i;
    if (count == m) {
      ans = j;
      break;
    }
  }
  if (ans == 0)
    cout << 0 << endl;
  else {
    cout << 5 << endl;
    for (int i = 0; i < 5; i++) cout << ans + i << " ";
  }
}

### Prompt

Create a solution in CPP for 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 main() {
  long long m;
  cin >> m;
  long long n = 1000000, count = 0, ans = 0;
  for (long long j = 5; j < n; j++) {
    count = 0;
    for (long long i = 5; j / i > 0; i = i * 5) count += j / i;
    if (count == m) {
      ans = j;
      break;
    }
  }
  if (ans == 0)
    cout << 0 << endl;
  else {
    cout << 5 << endl;
    for (int i = 0; i < 5; i++) cout << ans + i << " ";
  }
}

```