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(non_snake_case)]

use std::io::{stdin, Read};
use std::str::FromStr;
fn read_option<T: FromStr>() -> Option<T> {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok()
}
fn read<T: FromStr>() -> T {
    let opt = read_option();
    opt.expect("failed to parse token")
}

fn main() {
    let x: usize = read();
    for i in 1..180 {
        if 360 * i % x == 0 {
            println!("{}", 360 * i / x);
            return;
        }
    }
}