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 bisect
  import copy
  import decimal
  import fractions
  import heapq
  import itertools
  import math
  import random
  import sys
  from collections import Counter,deque,defaultdict
  from functools import lru_cache,reduce
  from heapq import heappush,heappop,heapify,heappushpop,_heappop_max,_heapify_max
  def _heappush_max(heap,item):
      heap.append(item)
      heapq._siftdown_max(heap, 0, len(heap)-1)
  def _heappushpop_max(heap, item):
      if heap and item < heap[0]:
          item, heap[0] = heap[0], item
          heapq._siftup_max(heap, 0)
      return item
  from math import gcd as GCD, modf
  read=sys.stdin.read
  readline=sys.stdin.readline
  readlines=sys.stdin.readlines
  
  t=int(readline())
  for _ in range(t):
      N,M=map(int,readline().split())
      M+=1
      flag=True
      ans=0
      for i in range(31,-1,-1):
          if not N>>i&1 and M>>i&1 and flag:
              ans^=1<<i
          if flag and N>>i&1 and not M>>i&1:
              flag=False
      print(ans)