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: from __future__ import division,print_function
  from heapq import*
  import sys
  le = sys.__stdin__.read().split("\n")[::-1]
  af=[]
  for zorg in range(int(le.pop())):
      n,m= list(map(int,le.pop().split()))
      #answer is x^n where x is smallest number  bigger than m with x&n==n
      #l=[n^k for k in range(m+1,m+n+1)]
      #r=m+1+l.index(min(l))
      if n==0:
          af.append(m+1)
      else:
          mb=0#maximal bit of n privé de m
          for i in range(30):
              if (2**i & n) and not(2**i & (m+1)):
                  mb=i
          r=0
          for i in range(30):
              if (2**i & n) or ((2**i & (m+1)) and i>=mb):
                  r+=2**i
          #return r,r^n,((m+1)|n)^n,((m+1)|n),r&n
          af.append(r^n)
  print("\n".join(map(str,af)))
  