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 io, os
  import math
  from heapq import *
  gcd = math.gcd
  sqrt = math.sqrt
  ceil = math.ceil
  # arr=list(map(int, input().split()))
  def strinp(testcases):
      k = 5
      if (testcases == -1 or testcases == 1):
          k = 1
      f = str(input())
      f = f[2:len(f) - k]
      return f
  def print(x, end='\n'):
      sys.stdout.write(str(x) + end)
  def ind(ch):
      return ord(ch)-ord("a")
  def db(n):
      k=list(bin(n).replace("0b", ""))
      f=len(k)
      rem=31-f
      k=(["0"]*rem)+k
      return k
  def bd(arr):
      p=1
      s=0
      for i in range(len(arr)-1,-1,-1):
          s+=(p*int(arr[i]))
          p*=2
      return s
  def main():
      t=int(input())
      for _ in range(t):
          arr=list(map(int, input().split()))
          n=arr[0]
          m=arr[1]
          if(n>m):
              print(0)
              continue
          bm=db(m)
          bn=db(n)
          ans=["0"]*31
          f=30
          for i in range(31):
              if(bm[i]=="1"):
                  f=i
                  break
          g=-1
          xor=[0]*31
          trig=False
          for i in range(31):
              if(bm[i]=="1" and bn[i]=="0"):
                  ans[i]="1"
              if(bm[i]=="0" and bn[i]=="0"):
                  g=i
              xor[i]=int(bn[i])^int(ans[i])
              if(xor[i]>int(bm[i])):
                  trig=True
                  break
          if(trig):
              c = bd(ans)
              print(c)
              continue
          ans[g]="1"
          for i in range(g+1,31):
              ans[i]="0"
          print(bd(ans))
  main()