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, os, io, heapq
  from sys import stdin, stdout
  from collections import Counter, deque, defaultdict
  from itertools import product
  import bisect
  
  
  # For fast i/o
  # def inp():
  #     return io.BytesIO(os.read(0,os.fstat(0).st_size)).readline().strip().decode()
  # input = inp
  
  # def println(v, end='\n'):
  #     stdout.write(str(v) + end)
  
  def getList(method=int):
      return list(map(method, input().split()))
  
  
  def getInt():
      return int(input())
  
  
  def getXor(n):
      r = n % 4
      if r == 0:
          return n
      if r == 1:
          return 1
      if r == 2:
          return n + 1
      return 0
  
  
  def solve():
      a, b = getList()
      x = getXor(a - 1)
  
      if x == b:
          return a
  
      if b ^ x != a:
          return a + 1
  
      return a + 2
  
  
  for _ in range(getInt()):
      print(solve())
  
  # x = 0
  # for i in range(20):
  #     x ^= i
  #     print(x)
  
  