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
  import sys
  import collections
  import bisect
  import time
  import random
  import itertools
  import string
  from collections import Counter,defaultdict
  def get_ints():return map(int, sys.stdin.readline().strip().split())
  def get_list():return list(map(int, sys.stdin.readline().strip().split()))
  def get_string():return sys.stdin.readline().strip()
  def computeXOR(n):
      if n % 4 == 0:
          return n
      if n % 4 == 1:
          return 1
      if n % 4 == 2:
          return n + 1
      return 0
  for t in range(int(input())):
      a,b=get_ints()
      ans=a
      curr_xor=computeXOR(a-1)
      if curr_xor==b:
          print(ans)
      else:
          no_req=b^curr_xor
          if no_req==a:
              print(ans+2)
          else:print(ans+1)