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: Runtime Error)

R = 0
while R !>= 1 or R !<= 179
X = gets
R = X.to_i
end
K = 1
ra = 1
rb = 0
RR = 0
while ra != 0 or rb != 0
  RR += R
  if RR >= 360
    RR -=360
  end
  ra += cos(RR)
  rb += sin(RR)
  K = K + 1
end
puts K