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: # ///////////////////////////////////////////////////////////////////////////
  # //////////////////// PYTHON IS THE BEST ////////////////////////
  # ///////////////////////////////////////////////////////////////////////////
  import sys,os,io
  from sys import stdin
  import math 
  from collections import defaultdict
  from heapq import heappush, heappop, heapify
  from bisect import bisect_left , bisect_right
  from io import BytesIO, IOBase
  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")
  
  
  
  
  alphabets = list('abcdefghijklmnopqrstuvwxyz')
  
  
  
  #for deep recursion__________________________________________-
  from types import GeneratorType
  def bootstrap(f, stack=[]):
      def wrappedfunc(*args, **kwargs):
          if stack:
              return f(*args, **kwargs)
          else:
              to = f(*args, **kwargs)
              while True:
                  if type(to) is GeneratorType:
                      stack.append(to)
                      to = next(to)
                  else:
                      stack.pop()
                      if not stack:
                          break
                      to = stack[-1].send(to)
              return to
  
      return wrappedfunc
  
  def ncr(n, r, p):  
      num = den = 1
      for i in range(r):
          num = (num * (n - i)) % p
          den = (den * (i + 1)) % p
      return (num * pow(den,p - 2, p)) % p
  
  def primeFactors(n): 
      l = []
      while n % 2 == 0: 
          l.append(2)
          n = n / 2
      for i in range(3,int(math.sqrt(n))+1,2): 
          while n % i== 0: 
              l.append(int(i))
              n = n / i 
      if n > 2: 
          l.append(n)
      # c = dict(Counter(l))
      return list(set(l))
      # return c
  
  def power(x, y, p) : 
  	res = 1
  	x = x % p 
  	if (x == 0) : 
  		return 0
  	while (y > 0) : 
  		if ((y & 1) == 1) : 
  			res = (res * x) % p 
  		y = y >> 1	 # y = y/2 
  		x = (x * x) % p 		
  	return res 
  
  #____________________GetPrimeFactors in log(n)________________________________________
  def sieveForSmallestPrimeFactor():
      MAXN = 100001
      spf = [0 for i in range(MAXN)]
      spf[1] = 1
      for i in range(2, MAXN):
          spf[i] = i
      for i in range(4, MAXN, 2):
          spf[i] = 2
      for i in range(3, math.ceil(math.sqrt(MAXN))):
          if (spf[i] == i):
              for j in range(i * i, MAXN, i): 
                  if (spf[j] == j):
                      spf[j] = i
      return spf
  def getPrimeFactorizationLOGN(x):
      spf = sieveForSmallestPrimeFactor()
      ret = list()
      while (x != 1):
          ret.append(spf[x])
          x = x // spf[x]  
      return ret
  #____________________________________________________________
  
  
  
  def SieveOfEratosthenes(n): 
      #time complexity = nlog(log(n))
      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
      return prime
  def si():
      return input()
  def divideCeil(n,x):
      if (n%x==0):
          return n//x
      return n//x+1
  def ii():
      return int(input())
  def li():
      return list(map(int,input().split()))
  
  
  # ///////////////////////////////////////////////////////////////////////////
  # //////////////////// DO NOT TOUCH BEFORE THIS LINE ////////////////////////
  # ///////////////////////////////////////////////////////////////////////////
  if(os.path.exists('input.txt')):
      sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w") 
  else:
      sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
      input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  
  def solve():
      n,m = li()
      if m<n:
          print(0)
          return 
      m+=1
      n = list(bin(n)[2:])
      m = list(bin(m)[2:])
      x = max(len(n),len(m))
      if len(n)<x:
          n  = ['0']*(x-len(n))+n
      else:
          m  = ['0']*(x-len(m))+m
      ans = []
      # print(n)
      # print(m)
  
      for i in range(len(n)):
          if n[i]==m[i]:
              ans.append('0')
          elif n[i]=='0' and m[i]=='1':
              ans.append('1')
          else:
              break 
      ans = ans + ['0']*(len(n)-len(ans))
      ans = ''.join(ans)
      print(int(ans,2))
  
  
  
  t = 1
  t = ii()
  for _ in range(t):
      solve()
  