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)

include Math
X = gets.to_i

# $count = 0
#
# def move(x, y, count)
#   theta = (PI / 2) + X * ( PI / 180 ) * $count
#   complex = Complex(1 - x, 0 - y) * Complex(cos(theta).to_c,sin(theta)to_c) + Complex(x, y)
#   p complex
#   x = complex.real
#   y = complex.imag
#   return $count if x.to_i == 0 && y.to_i == 0
#   $count += 1
#   move(x, y, $count)
# end
#
# move(0, 0, $count)
# p $count
#
# #
# # theta = (PI / 2) + X * ( PI / 180 ) * $count
# # x = (1-x) * cos(theta) + (y * sin(theta)) + x
# # p x.to_i
# # y = (1-x) * sin(theta) - y * cos(theta) + y

result = 0
for i in 1..360
  result = i if i * X == 360
end
puts result