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 collections
  # import random
  # import math
  # from collections import defaultdict
  # import itertools
  # from sys import stdin, stdout
  import sys
  # import operator
  # from decimal import Decimal
  
  # sys.setrecursionlimit(10**6)
  
  p2D = lambda x: print(*x, sep="\n")
  def II(): return int(sys.stdin.buffer.readline())
  def MI(): return map(int, sys.stdin.buffer.readline().split())
  def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
  def LLI(rows_number): return [LI() for _ in range(rows_number)]
  def BI(): return sys.stdin.buffer.readline().rstrip()
  def SI(): return sys.stdin.buffer.readline().rstrip().decode()
  def li(): return [int(i) for i in input().split()]
  def lli(rows): return [li() for _ in range(rows)]
  def si(): return input()
  def ii(): return int(input())
  def ins(): return input().split()
  
  
  def solve():
      n, m = LI()
      if m<n:
          return 0
      elif n == m:
          if n%2 == 0:
              return 1
          else:
              c = 0
              while n>0:
                  if n%2 ==0:
                      ans = c
                      break
                  n//=2
                  c+=1
  
              return 2**c
      else:
          nb = bin(n).lstrip('0b')
          mb = bin(m).lstrip('0b')
  
          nb ='0'*(len(mb)-len(nb))+nb
          # print(nb)
          # print(mb)
          l = len(nb)
          sb = ''
          for i in range(l):
              if nb[i]<=mb[i]:
                  sb+= str(int(nb[i])^int(mb[i]))
              else:
                  sb+= '0'*(l-i)
                  return int(sb,2)
  
          else:
              c = 0
              while m > 0:
                  if n % 2 == 0 and m%2 == 0:
                      ans = sb[:l-c-1]+ '1' + '0'*(c)
                      return int(ans,2)
                  n //= 2
                  m //= 2
                  c += 1
              else:
                  return 2**c
  
  
  def main():
      for _ in range(II()):
          sys.stdout.write(str(solve()) + "\n")
  
  
      #     z += str(ans) + '\n'
      # print(len(ans), ' '.join(map(str, ans)), sep='\n')
      # stdout.write(z)
  
  
  # for interactive problems
  # print("? {} {}".format(l,m), flush=True)
  # or print this after each print statement
  # sys.stdout.flush()
  
  
  if __name__ == "__main__":
      main()