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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(unused, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};
fn main() {
    input! {//
        n:usize,
        // mut a:[i64;n],
    }
    // a = a.iter_mut().map(|&mut x| x % 2).collect::<Vec<_>>();
    // dbg!(&a);
    let mut ans = 0;
    let mut angle = 0;
    for i in 1..33333 {
        angle += n;
        if angle % 360 == 0 {
            println!("{}", i);
            return;
        }
    }
}