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 Ruby (2.7.1) (judge verdict: Wrong Answer)

### SNIPPET
  # n = gets.split.map(&:to_i)
  # array = n.times.map { gets.split.map(&:to_i) }
  # [].all?(&:even?)
  # a = [*1..m].repeated_combination(n).to_a
  # [1,2,3,4,5].select { |num| num.even? }  # => [2, 4]
  # ["a","a","b"].group_by(&:itself).map{|k,v| [k, v.count]}.to_h #=> {"a"=>2, "b"=>1}
  # 切り捨て: .floor(2).to_f ,切り上げ： .ceil(2).to_f,四捨五入: round(2)
  # 3.upto(6) do |i|, 6.downto(3) do |i|
  # 公約数125.gcd(100)、公倍数125.lcm(100)
  # PI = Math::PI
  # 高さ = a * Math.sin(w / 180.0 * Math::PI), 底辺 = a * Math.cos(w / 180.0 * Math::PI)
  # def chmax(a, b) a > b ? a : b end
  # INF = Float::INFINITY
  # def chmin(a, b) a < b ? a : b end

x = gets.to_i
if x > 90
  x = 180 - x
  puts (360 / x) / 2
else
  puts 360 / x
end