#include <bits/stdc++.h>
using namespace std;
int n;
int calc(int x) {
  int a = 0;
  for (int i = 5; i <= x; i *= 5) a += x / i;
  return a;
}
int judge(int x) {
  int l = 5, r = 500000, mid, res = -1;
  while (l <= r) {
    mid = (l + r) / 2;
    if (calc(mid) >= x)
      res = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  int l = judge(n), r = judge(n + 1);
  cout << r - l << endl;
  for (int i = l; i <= r - 1; i++) cout << i << ' ';
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 n;
int calc(int x) {
  int a = 0;
  for (int i = 5; i <= x; i *= 5) a += x / i;
  return a;
}
int judge(int x) {
  int l = 5, r = 500000, mid, res = -1;
  while (l <= r) {
    mid = (l + r) / 2;
    if (calc(mid) >= x)
      res = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  int l = judge(n), r = judge(n + 1);
  cout << r - l << endl;
  for (int i = l; i <= r - 1; i++) cout << i << ' ';
  return 0;
}

```