#include <bits/stdc++.h>
using namespace std;
int fun(int n) {
  int ans = 0;
  while (n) {
    ans += n / 5;
    n = n / 5;
  }
  return ans;
}
int bsearh(int x, int y, int v) {
  int ans = 0;
  int mid = 0;
  while (x <= y) {
    mid = (x + y) / 2;
    int cnt = fun(mid);
    if (cnt < v)
      x = x + 1;
    else {
      y = mid - 1;
      ans = mid;
    }
  }
  return ans;
}
int main() {
  int m;
  while (cin >> m) {
    int i, x = 1, y = 1000000000;
    int L = bsearh(x, y, m);
    int R = bsearh(x, y, m + 1);
    if (L == 0 || fun(R - 1) != m) {
      cout << 0 << endl;
    } else {
      cout << R - L << endl;
      for (i = L; i <= R - 1; i++) {
        cout << i;
        if (i != R - 1)
          cout << " ";
        else
          cout << endl;
      }
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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;
int fun(int n) {
  int ans = 0;
  while (n) {
    ans += n / 5;
    n = n / 5;
  }
  return ans;
}
int bsearh(int x, int y, int v) {
  int ans = 0;
  int mid = 0;
  while (x <= y) {
    mid = (x + y) / 2;
    int cnt = fun(mid);
    if (cnt < v)
      x = x + 1;
    else {
      y = mid - 1;
      ans = mid;
    }
  }
  return ans;
}
int main() {
  int m;
  while (cin >> m) {
    int i, x = 1, y = 1000000000;
    int L = bsearh(x, y, m);
    int R = bsearh(x, y, m + 1);
    if (L == 0 || fun(R - 1) != m) {
      cout << 0 << endl;
    } else {
      cout << R - L << endl;
      for (i = L; i <= R - 1; i++) {
        cout << i;
        if (i != R - 1)
          cout << " ";
        else
          cout << endl;
      }
    }
  }
  return 0;
}

```