description: You are given two integers n and m. Find the \operatorname{MEX} of the sequence n ⊕ 0, n ⊕ 1, …, n ⊕ m. Here, ⊕ is the [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
  
  \operatorname{MEX} of the sequence of non-negative integers is the smallest non-negative integer that doesn't appear in this sequence. For example, \operatorname{MEX}(0, 1, 2, 4) = 3, and \operatorname{MEX}(1, 2021) = 0. 
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 30 000) — the number of test cases.
  
  The first and only line of each test case contains two integers n and m (0 ≤ n, m ≤ 10^9).
  
  Output
  
  For each test case, print a single integer — the answer to the problem.
  
  Example
  
  Input
  
  
  5
  3 5
  4 6
  3 2
  69 696
  123456 654321
  
  
  Output
  
  
  4
  3
  0
  640
  530866
  
  Note
  
  In the first test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, 3 ⊕ 3, 3 ⊕ 4, 3 ⊕ 5, or 3, 2, 1, 0, 7, 6. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 4.
  
  In the second test case, the sequence is 4 ⊕ 0, 4 ⊕ 1, 4 ⊕ 2, 4 ⊕ 3, 4 ⊕ 4, 4 ⊕ 5, 4 ⊕ 6, or 4, 5, 6, 7, 0, 1, 2. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 3.
  
  In the third test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, or 3, 2, 1. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 0.
solution: import sys
  import math
  
  if sys.subversion[0] == "PyPy":
      import io, atexit
  
      sys.stdout = io.BytesIO()
      atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
  
      sys.stdin = io.BytesIO(sys.stdin.read())
      input = lambda: sys.stdin.readline().rstrip()
  
  t = int(raw_input())
  
  # python codesforces210710.py < input.txt
  for zz in range(t):
      #n = int(raw_input())
      str2 = raw_input()
      n, m = str2.split(' ')
      n = int(n)
      m = int(m) + 1
      linen = [int(x) for x in bin(n)[2:]]
      lima = [int(x) for x in bin(m)[2:]]
      ans = 0
      fl = []
  
      bas = []
  
  
      if len(linen) <= len(lima):
          if len(linen) < len(lima):
              bas = lima[0:len(lima)-len(linen)]
              #print(''.join(map(str, likay[len(likay)-1:len(linen)-1:-1])))
              for i in range(len(lima) - len(linen)):
                  lima.pop(0)
          for i in range(len(lima)):
              if lima[i] == 0 and linen[i] == 1:
                  lili = [0] * (len(lima) - i)
                  fl = fl + lili
                  break
              else:
                  fl.append(lima[i] - linen[i])
          fl = bas + fl
          if len(fl) > 0:
              ans = int(''.join(map(str, fl)), 2)
  
  
      print(ans)
  