You are given two very large numbers made up of 1 or 0 only. You have to find the digit by digit XOR of the two numbers, i.e., the i-th digit of the answer is 1 if and only if the i-th digit of the two given numbers differ otherwise the i-th digit of the answer is 0. The number of digits in both the numbers is same.

Input:

First line contains the number of test cases T.

Each test case is represented by two lines representing the two numbers. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same.

Output:

For each test case print the corresponding answer.

Constraints:

1 ≤ T ≤ 100

1 ≤ Number of digits in a number ≤ 10^4

SAMPLE INPUT
2
01110
01100
010
111

SAMPLE OUTPUT
00010
101
'''
# Read input from stdin and provide input before running code

name = raw_input('What is your name?\n')
print 'Hi, %s.' % name
'''

T = int(raw_input(""))

list_NUM=[]

for a in range(0,2*T):
	list_NUM.append(raw_input(""))
	
list_res=[]

counter=0
for k in range(0,T):
	listr=[]
	for c in range(0,len(list_NUM[counter])):
		if(list_NUM[counter][c] != list_NUM[counter+1][c]):
			listr.append("1")
		else:
			listr.append("0")
			
	list_res.append(listr)
	counter = counter + 2;
	
for j in range(0,T):
	print ''.join(list_res[j])