#include <bits/stdc++.h>
bool get(int x, int m) {
  int count = 0;
  int t = 5;
  while (t <= x) {
    count += x / t;
    t *= 5;
  }
  return count == m;
}
int main() {
  int m;
  std::vector<int> ans;
  int count = 0;
  std::cin >> m;
  for (int i = 1; i <= m * 10; ++i) {
    if (get(i, m)) {
      count++;
      ans.push_back(i);
    }
  }
  std::cout << count << std::endl;
  for (int x : ans) std::cout << x << ' ';
  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>
bool get(int x, int m) {
  int count = 0;
  int t = 5;
  while (t <= x) {
    count += x / t;
    t *= 5;
  }
  return count == m;
}
int main() {
  int m;
  std::vector<int> ans;
  int count = 0;
  std::cin >> m;
  for (int i = 1; i <= m * 10; ++i) {
    if (get(i, m)) {
      count++;
      ans.push_back(i);
    }
  }
  std::cout << count << std::endl;
  for (int x : ans) std::cout << x << ' ';
  return 0;
}

```