#include <bits/stdc++.h>
using namespace std;
bool vis[100005];
int main() {
  int n;
  while (cin >> n) {
    int i, tag = 0, num = 0;
    for (i = 5;; i += 5) {
      int x = i;
      while (x % 5 == 0 && x) {
        num++;
        x /= 5;
      }
      if (num == n) {
        tag = 1;
        break;
      } else if (num > n)
        break;
    }
    if (tag) {
      cout << "5" << endl;
      for (int p = 0; p < 5; p++) cout << i + p << " ";
      cout << endl;
    } else
      cout << "0" << endl;
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
bool vis[100005];
int main() {
  int n;
  while (cin >> n) {
    int i, tag = 0, num = 0;
    for (i = 5;; i += 5) {
      int x = i;
      while (x % 5 == 0 && x) {
        num++;
        x /= 5;
      }
      if (num == n) {
        tag = 1;
        break;
      } else if (num > n)
        break;
    }
    if (tag) {
      cout << "5" << endl;
      for (int p = 0; p < 5; p++) cout << i + p << " ";
      cout << endl;
    } else
      cout << "0" << endl;
  }
  return 0;
}

```