#include<bits/stdc++.h>
using namespace std;

int main()
{
  int x,k;
  cin>>x;
  for(k=1;;k++)
    if((k*x)%360==0){
     cout<<k<<endl;
     return 0;
    }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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<bits/stdc++.h>
using namespace std;

int main()
{
  int x,k;
  cin>>x;
  for(k=1;;k++)
    if((k*x)%360==0){
     cout<<k<<endl;
     return 0;
    }
}

```