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 os
  from io import BytesIO
  
  # pypy2
  sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
  
  f = sys.stdin
  
  line = lambda: f.readline().strip('\r\n').split()
  
  def write(w):
      sys.stdout.write(w)
      sys.stdout.write("\n")
  
  def solve():
      if M <= N:
          return("0")
      
      res = 0
      m_bits = M.bit_length()
      n_bits = N.bit_length()
      for i in range(m_bits-n_bits):
          if M&(1<<(n_bits+i)):
              res |= 1<<(n_bits+i)
      for i in range(n_bits-1,-1,-1):
          m = M&(1<<i)
          n = N&(1<<i)
          if m and not n:
              res |= 1<<i
          elif not m and n:
              break
          
      return str(res)
  
  T = int(line()[0])
  for test in range(1,T+1):
      N,M = map(int,line())
      M += 1
      
      write(solve())
      
  f.close()