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 math
  from sys import stdin, stdout
  # from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
  from collections import defaultdict as df, deque
  from heapq import merge, heapify, heappop, heappush, nsmallest
  from bisect import bisect_left as bl, bisect_right as br, bisect
  mod = pow(10, 9) + 7
  mod2 = 998244353
  S1 = 'abcdefghijklmnopqrstuvwxyz'
  S2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
  def SieveOfEratosthenes(n):
      l=[]
      prime = [True for i in range(n+1)]
      p = 2
      while (p * p <= n):
          if (prime[p] == True):
              for i in range(p * p, n+1, p):
                  prime[i] = False
          p += 1
      for p in range(2, n+1):
          if prime[p]:
              l.append(p)
      return l
  def isprime(x):
      if x<=1: return False
      if x in (2, 3): return True
      if x%2 == 0: return False
      for i in range(3, int(sqrt(x))+1, 2):
          if x%i == 0: return False
      return True
  for _ in range(int(input())):
      n,m=map(int,input().split())
      if n>m:
          print(0)
      else:
          # if n==m:
          #     print(1)
          # else:
          if n==0:
              print(m+1)
              # elif n==1:
              #     if m%2==0:
              #         print(m)
              #     else:
              #         print(m+1)
              # elif n==2:
              #     if (m-2)%4==0:
              #         print((n^(m-4+1))+1)
              #     else:
              #         print(n^(m-4+1))
          else:
              m=m+1
              mex=0
              cnt=32
              while(cnt>-1):
                  a=n&(1<<cnt)
                  b=m&(1<<cnt)
                  if a<b:
                      mex+=b
                  if b<a:
                      break
                  cnt-=1
              print(mex)
  
          
  
  
              
          
  
  
  # n,k=map(int,input().split())
  # a=list(map(int,input().split()))
  
      # max1=-1
      # for i in range(1,len(a)):
      #     max1=max(max1,a[i]*a[i-1])
      # print(max1)
  
  