#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  long long int n, i, c, k = 0, z = 0, cnt = 0, a[100010];
  cin >> n;
  while (1) {
    if (z > n) break;
    c = k;
    while (c % 5 == 0) {
      if (c == 0) break;
      c /= 5;
      z++;
    }
    if (z == n) a[cnt++] = k;
    k++;
  }
  cout << cnt << endl;
  for (i = 0; i < cnt; i++) cout << a[i] << " ";
  return 0;
}

### 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() {
  ios::sync_with_stdio(false);
  long long int n, i, c, k = 0, z = 0, cnt = 0, a[100010];
  cin >> n;
  while (1) {
    if (z > n) break;
    c = k;
    while (c % 5 == 0) {
      if (c == 0) break;
      c /= 5;
      z++;
    }
    if (z == n) a[cnt++] = k;
    k++;
  }
  cout << cnt << endl;
  for (i = 0; i < cnt; i++) cout << a[i] << " ";
  return 0;
}

```