description: Alice gave Bob two integers a and b (a > 0 and b ≥ 0). Being a curious boy, Bob wrote down an array of non-negative integers with \operatorname{MEX} value of all elements equal to a and \operatorname{XOR} value of all elements equal to b.
  
  What is the shortest possible length of the array Bob wrote?
  
  Recall that the \operatorname{MEX} (Minimum EXcluded) of an array is the minimum non-negative integer that does not belong to the array and the \operatorname{XOR} of an array is the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all the elements of the array.
  
  Input
  
  The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 5 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
  
  The only line of each test case contains two integers a and b (1 ≤ a ≤ 3 ⋅ 10^5; 0 ≤ b ≤ 3 ⋅ 10^5) — the \operatorname{MEX} and \operatorname{XOR} of the array, respectively.
  
  Output
  
  For each test case, output one (positive) integer — the length of the shortest array with \operatorname{MEX} a and \operatorname{XOR} b. We can show that such an array always exists.
  
  Example
  
  Input
  
  
  5
  1 1
  2 1
  2 0
  1 10000
  2 10000
  
  
  Output
  
  
  3
  2
  3
  2
  3
  
  Note
  
  In the first test case, one of the shortest arrays with \operatorname{MEX} 1 and \operatorname{XOR} 1 is [0, 2020, 2021].
  
  In the second test case, one of the shortest arrays with \operatorname{MEX} 2 and \operatorname{XOR} 1 is [0, 1].
  
  It can be shown that these arrays are the shortest arrays possible.
solution: import math,sys
  #from itertools import permutations, combinations;import heapq,random;
  from collections import defaultdict,deque
  import bisect as bi
  def yes():print('YES')
  def no():print('NO')
  #sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
  def I():return (int(sys.stdin.readline()))
  def In():return(map(int,sys.stdin.readline().split()))
  def Sn():return sys.stdin.readline().strip()
  #sys.setrecursionlimit(1500)
  def dict(a):
      d={} 
      for x in a:
          if d.get(x,-1)!=-1:
              d[x]+=1
          else:
              d[x]=1
      return d
  def find_gt(a, x):
      'Find leftmost value greater than x'
      i = bi.bisect_left(a, x)
      if i != len(a):
          return i
      else:            
          return -1
  def main(dp):
      try:
          a,b=In()
          xt=dp[a-1]^b
          if xt:
              if xt==a:
                  a+=1
              print(a+1)
          else:
              print(a)
      except:
          pass
          
  M = 998244353
  P = 1000000007
   
  if __name__ == '__main__':
      dp=[0]
      for i in range(1,300001):
          dp.append(dp[-1]^i)
      for _ in range(I()):main(dp)
      #for _ in range(1):main()
  #End#
  
  #        ******************* All The Best *******************   #