#include <bits/stdc++.h>
using namespace std;
const int N = 5e2 + 10;
int ans, mac, n;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  while (ans < n) {
    mac += 5;
    int x = mac;
    while (x % 5 == 0) {
      x /= 5;
      ans++;
    }
  }
  if (ans == n) {
    cout << 5 << '\n';
    for (int i = mac; i < mac + 5; ++i) {
      cout << i << ' ';
    }
    cout << '\n';
    return 0;
  }
  cout << 0 << '\n';
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int N = 5e2 + 10;
int ans, mac, n;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  while (ans < n) {
    mac += 5;
    int x = mac;
    while (x % 5 == 0) {
      x /= 5;
      ans++;
    }
  }
  if (ans == n) {
    cout << 5 << '\n';
    for (int i = mac; i < mac + 5; ++i) {
      cout << i << ' ';
    }
    cout << '\n';
    return 0;
  }
  cout << 0 << '\n';
  return 0;
}

```