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: def process(n, m):
      if n > m:
          return [0, 0]
      n1 = bin(n)[2:]
      m1 = bin(m)[2:]
      n1 = n1.zfill(len(m1))
      n1 = n1[::-1]
      m1 = m1[::-1]
    #  print(n1)
     # print(m1)
     # g = m-n
     # if (n^(g+1))==n+g+1:
     #     return g+1
      answer = 0
      is_good = None
      for i in range(len(m1)):
          if n1[i]=='1' and m1[i]=='0':
              is_good = i
      if is_good is None:
          first_zero = None
          for i in range(len(m1)):
              if m1[i]=='0':
                  first_zero = i
                  break
          if first_zero is None:
              return [2**(len(m1)), 0]
          answer = 2**(first_zero)
          for i in range(first_zero+1, len(m1)):
              if m1[i]=='1' and n1[i]=='0':
                  answer+=(2**i)
          return [answer, 0]
      #so this seems to be fine
      
      for i in range(len(m1)):
          if i > is_good and n1[i]=='0' and m1[i]=='1':
              answer+=(2**i)
      return [answer, 1]
  
  t = int(input())
  for i in range(t):
      n, m = [int(x) for x in input().split()]
      print(process(n, m)[0])