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)

use proconio::input;

// 最大公約数 (あえて match を使ってみる)
fn gcd(a: usize, b: usize) -> usize {
    match b {
        0 => a,
        _ => gcd(b, a % b)
    }
}

// 最小公倍数
fn lcm(a: usize, b: usize) -> usize {
    a * b / gcd(a, b)
}

fn main() {
	input!(x: usize);

	let lcm = lcm(360, x);
	println!("{}", lcm / x);
}