#include <iostream>
using namespace std;
int main(){
  int x;
  cin >> x;
  int i=1;
  while((360*i)%x!=0){
    i++;
  }
  cout << (360*i)/x;
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

Takahashi is standing on a two-dimensional plane, facing north. Find the minimum positive integer K such that Takahashi will be at the starting position again after he does the following action K times:

* Go one meter in the direction he is facing. Then, turn X degrees counter-clockwise.

Constraints

* 1 \leq X \leq 179
* X is an integer.

Input

Input is given from Standard Input in the following format:


X


Output

Print the number of times Takahashi will do the action before he is at the starting position again.

Examples

Input

90


Output

4


Input

1


Output

360

### Response

```cpp
#include <iostream>
using namespace std;
int main(){
  int x;
  cin >> x;
  int i=1;
  while((360*i)%x!=0){
    i++;
  }
  cout << (360*i)/x;
  return 0;
}

```