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: from collections import defaultdict as dd
  import math
  import sys
  input=sys.stdin.readline
  def nn():
      return int(input())
   
  def li():
      return list(input())
   
  def mi():
      return map(int, input().split())
   
  def lm():
      return list(map(int, input().split()))
   
   
   
  def solve():
      n,m = mi()
      
      if n==0 and m==0:
          print(1)
          return
      if n>m:
          print(0)
          return
      exp = 40
      
      while 2**exp > m:
          exp-=1
  
      val = 2**(exp+1)-1
      if m==val:
          print(m+1)
          return
      while exp>=0:
          temp = n
          #print(val,n, (temp>>exp)%2)
          
          if (val - 2**exp) > m and ((temp>>exp)%2)==0:
              val -= 2**exp
          exp-=1
          
      #print(val)
      print(n^val)
      
  
  q=nn()
  for _ in range(q):
      solve()