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 <bits/stdc++.h>
using namespace std;
 
typedef long long LL;
typedef vector < int > VI;
typedef vector < VI > VVI;
typedef pair < int , int > PII;
 
#define ALL(x) x.begin(),x.end()
#define PB push_back
#define MP make_pair
#define SZ(x) x.size()
#define SET(a , b) memset(a , b , sizeof(a))
#define REP(i, n) for(int i = 0 ; i < n ; i++)
#define FOR(i, a, b) for(int i = a ; i <= b ; i++)
#define FORD(i, a, b) for(int i = a ; i >= b ; i--)
#define BOOST ios_base::sync_with_stdio(0);
#define TCSOLVE() int tests; cin >> tests; REP(i, tests) solve();
 
const int N = 2e5 + 5, M = 2e5 + 5, mx = 1e9, LN = 20, mod = 1e9 + 7;



int main(){
	int x; cin >> x;

	FOR(i, 2, 360){
		if((x * i) % 360 == 0){
			cout << i << endl;
			return 0;
		}
	}
}