# Task
 You are a magician. You're going to perform a trick.

 You have `b` black marbles and `w` white marbles in your magic hat, and an infinite supply of black and white marbles that you can pull out of nowhere. 
 
 You ask your audience to repeatedly remove a pair of marbles from your hat and, for each pair removed, you add one marble to the hat according to the following rule until there is only 1 marble left. 

 If the marbles of the pair that is removed are of the same color, you add a white marble to the hat. Otherwise, if one is black and one is white, you add a black marble.
 
 Given the initial number of black and white marbles in your hat, your trick is to predict the color of the last marble.
 
 Note: A magician may confuse your eyes, but not your mind ;-)

# Input/Output


 - `[input]` integer `b`

  Initial number of black marbles in the hat. 
  
  `1 <= b <= 10^9`


 - `[input]` integer `w`

  Initial number of white marbles in the hat. 
  
  `1 <= w <= 10^9`


 - `[output]` a string

  `"Black"` or `"White"` if you can safely predict the color of the last marble. If not, return `"Unsure"`.

Source: codewars | Difficulty: EASY
Tags: Ad-hoc

Starter code: def not_so_random(b,w):
	

Solutions:
def not_so_random(b, w):
	return ['White', 'Black'][b % 2]


def not_so_random(b, w):
	if b % 2 == 0 and w % 2 == 0:
		return 'White'
	elif b % 2 == 1 and w % 2 == 0:
		return 'Black'
	elif b % 2 == 1 and w % 2 == 1:
		return 'Black'
	elif b % 2 == 0 and w % 2 == 1:
		return 'White'


def not_so_random(b, _):
	return 'Black' if b & 1 else 'White'


not_so_random = lambda b, _: ['White', 'Black'][b & 1]


def not_so_random(b, w):
	return 'Black' if b & 1 else 'White'