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
'''
nums=[] 
i=0
count=0
prev_count=0
T = int(raw_input())
if T>=1 and T<=100:
  for num in range(T*2):
    nums.append(str(raw_input()))
  for num in range(len(nums)):
    prev_count=len(str(nums[num-1]))
    count=len(str(nums[num]))
    #print str(nums[num])
    
    if num%2==1 and count==prev_count and count>=1 and count<=100000 and prev_count>=1 and prev_count<=100000 :
       #print "done"
       tt=int(str(nums[num-1]),2) ^ int(str(nums[num]),2)
       #print tt
       kk=str(bin(tt))
       ll=(kk.split("b")[-1])
       #print ll
       cc=''
       for i in range(count-len(ll)):
       	  cc=cc+"0"
       print cc+ll
    if num%2==1:
       #print prev_count,count
       prev_count=count=0