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 atexit, io, sys, collections, math, heapq, fractions,copy, os, functools 
  import sys
  import random
  import collections
  from io import BytesIO, IOBase
   
  #####################################  python 3 START
  BUFSIZE = 8192
   
  
  class FastIO(IOBase):
      newlines = 0
   
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
   
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
   
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
   
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
   
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  def input():
      return raw_input() 
  
  
  def f(n,m):
      if n > m:
          return 0
      # now n <= m
  
      bn = list(map(int,bin(n)[2:]))
      bm = list(map(int,bin(m)[2:]))
      
      if len(bn) < len(bm): bn = [0 for ii in range(len(bm) - len(bn))] + bn
  
      if all([e == 1 for e in bm]):
          return 1 << len(bn)
  
      q = collections.deque([])
      ans = 0
      for i in range(len(bn)):
          if bn[i] == 0 and bm[i] == 0:
              q.append((1  << (len(bn) -1- i), len(bn) -1- i))
          elif bn[i] == 1 and bm[i] == 0:
              return ans
          elif bn[i] == 0 and bm[i] == 1:
              ans += 1 << (len(bn) - 1 - i)
          elif  bn[i] == 1 and bm[i] == 1:
              ans += 0
      u,idx = q.pop()
      ans += u 
      for ii in range(idx):
          if (ans >> ii) & 1:
              ans -= 1 << ii
      return ans
      print ()
      print (list(map(int,bin(ans)[2:])))
      print ((ans-2) ^ n > m)
      print (ans ^ n > m)
      return ans
      ans = 0
      for ii in range(0, m + 1):
          if ii ^ n <= m:
              continue
          return ii
      return n+1
  
  for _ in range(int(input())):
      n,m = map(int, input().split())    
      print (f(n,m))