#include <bits/stdc++.h>
using namespace std;
vector<int> e;
int solve(int x) {
  int res = 0;
  while (x % 5 == 0) {
    x /= 5;
    res++;
  }
  return res;
}
int main() {
  int n;
  cin >> n;
  int ans = 0, i = 5;
  while (ans <= n) {
    ans += solve(i);
    if (ans == n) e.push_back(i);
    i++;
  }
  cout << e.size() << endl;
  for (int i = 0; i < (int)e.size(); i++) {
    cout << e[i];
    if (i == (int)e.size() - 1)
      cout << endl;
    else
      cout << " ";
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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;
vector<int> e;
int solve(int x) {
  int res = 0;
  while (x % 5 == 0) {
    x /= 5;
    res++;
  }
  return res;
}
int main() {
  int n;
  cin >> n;
  int ans = 0, i = 5;
  while (ans <= n) {
    ans += solve(i);
    if (ans == n) e.push_back(i);
    i++;
  }
  cout << e.size() << endl;
  for (int i = 0; i < (int)e.size(); i++) {
    cout << e[i];
    if (i == (int)e.size() - 1)
      cout << endl;
    else
      cout << " ";
  }
  return 0;
}

```