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 solve(n,m):
      if n > m:
          return 0
      nbin = '{0:032b}'.format(n)
      mbin = '{0:032b}'.format(m)
  
      used= []
      free = []
      for i in reversed(range(len(mbin))):
          if nbin[31-i] == "1":
              used.append(2**i)
          else:
              free.append(2**i)
      # print(sum(used))
      left = m-n+1
      tmp = 0
      tot_sum = sum(free)
      for i in range(len(free)):
          if tot_sum - free[i] + tmp < left:
              # print(free[i])
              tmp += free[i]
          tot_sum -= free[i]
      return tmp
  
  import os
  import io
  # import time
  # a=time.time()
  if __name__ == "__main__":
      input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
  
  
      T = int(input().decode().strip())
      for t in range(T):
          n,k = [int(x) for x in input().decode().strip().split(" ")]
          res = solve(n,k)
          print(res)