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 math
  import random
  from queue import PriorityQueue as PQ
  from bisect import bisect_left as BSL
  from bisect import bisect_right as BSR
  from collections import OrderedDict as OD
  from collections import Counter
  from itertools import permutations
  # mod = 998244353
  mod = 1000000007
  MOD = mod
  # sys.setrecursionlimit(1000000)
  try:
      sys.stdin = open("actext.txt", "r")
      OPENFILE = 1
  except:
      pass
   
  def get_ints():
      return map(int,input().split())
   
  def palindrome(s):
      mid = len(s)//2
      for i in range(mid):
          if(s[i]!=s[len(s)-i-1]):
              return False
      return True
  def check(i,n):
      if(0<=i<n):
          return True
      else:
          return False
  
  # -----------------------------------------------------------------------------------------
  
  t = int(input())
  for tt in range(t):
      n,k = get_ints()
      n = '{:032b}'.format(n)
      k = '{:032b}'.format(k+1)
      s = ''
      for i in range(32):
          if(k[i]=='1' and n[i]=='1'):
              s+='0'
          elif(k[i]=='1'):
              s+='1'
          elif(n[i]=='1'):
              break
          else:
              s+='0'
      s = s+'0'*(32-len(s))
      print(int(s,2))
          
  
  
  