output_description: Print the number of times Takahashi will do the action before he is at the
  starting position again.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      X
attempt: import math
  
  turn_angle = int(input())
  
  current_angle = 90
  position_dict = {"x": 0, "y": 0}
  move_one = {"x": 0, "y": 1}
  num_actions = 1
  
  
  def change_angle(turn_angle, current_angle):
      current_angle = current_angle + turn_angle
      if current_angle > 360:
          current_angle = current_angle - 360
      return current_angle
  
  
  def resolve_direction(current_angle):
      radian_angle = (current_angle * math.pi) / 180
      move_dict = {
          "x": round(math.cos(radian_angle), 3),
          "y": round(math.sin(radian_angle), 3),
      }
      return move_dict
  
  
  def move_skater(position_dict, move_dict):
      for key, value in position_dict.items():
          position_dict[key] = round((value + move_dict.get(key, 0)), 3)
      return position_dict
  
  
  position_dict = move_skater(position_dict, move_one)
  current_angle = change_angle(turn_angle, current_angle)
  
  
  origin_dict = {"x": 0.0, "y": 0.0}
  
  while position_dict != origin_dict:
      move_dict = resolve_direction(current_angle)
      position_dict = move_skater(position_dict, move_dict)
      current_angle = change_angle(turn_angle, current_angle)
      num_actions = num_actions + 1
  
  print(num_actions)
  
problem_description: 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.
samples: [{"input": "90", "output": "4\n    \n\nTakahashi's path is a square.\n\n* * *"}, {"input": "1", "output": "360"}]