Problem:
Score : 200 points

Problem Statement

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.

Sample Input 1

90

Sample Output 1

4

Takahashi's path is a square.

Sample Input 2

1

Sample Output 2

360

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>

#define min(lhs,rhs) ((lhs)<(rhs)?(lhs):(rhs))

#define max(lhs,rhs) ((lhs)>(rhs)?(lhs):(rhs))

static inline int gcd(int a,int b){
    int r,x = min(a,b),y = max(a,b);
    while(x){
        r = y;
        r %= x;
        y = x;
        x = r;
    }
    return y;
}

static inline int lcm(int a,int b){
    return a * (b / gcd(a,b));
}

int main(void){
    int x;
    scanf("%d",&x);
    printf("%d",lcm(360,x)/x);
}