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 traceback
  import os
  import sys
  from io import BytesIO, IOBase
  
  BUFSIZE = 8192
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  sys.stdin = IOWrapper(sys.stdin)
  # sys.stdout = IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  #####################################################################
  
  import math
  from collections import defaultdict as dfd, Counter, deque
  from functools import lru_cache
  from itertools import accumulate
  from heapq import *
  from bisect import *
  # from sortedcontainers import SortedList
  
  def geti():
      return int(input())
  
  def gets():
      return input()
  
  def getil():
      return list(map(int, input().split()))
  
  def getsl():
      return input().split()
  
  def getinps(s):
      inps = s.split()
      m = {'i': geti, 's': gets, 'il': getil, 'sl': getsl}
      if len(inps) == 1: return m[s]()
      return [m[k]() for k in inps]
  
  def get2d(nrows, ncols, n=0):
      return [[n] * ncols for r in range(nrows)]
  
  def get_acc(a):
      return list(accumulate(a))
  
  def get_ncr(n, r, mod=10**9+7):
      num = den = 1
      r = min(r, n-r)
      for i in range(r):
          num = (num * (n - i)) % mod
          den = (den * (i + 1)) % mod
      return num * pow(den, mod - 2, mod) % mod
  
  def get_nc2(n):
      return n * (n-1) // 2
      
  def get_npr(n, r):
      if n < r: return 0
      return math.factorial(n) // math.factorial(r) 
  
  def put(a):
      return ' '.join(map(str, a))
  
  # sys.stdin = open('input.txt', 'r')
  # sys.stdout = open('output.txt', 'w')
  
  # while y:
  #     x, y = y, x % y
  
  ALPHABETS = 'abcdefghijklmnopqrstuvwxyz'
  VOWELS = 'aeiou'
  CONSONANTS = 'bcdfghjklmnpqrstvwxyz'
  
  inf = float('inf')
  mod = 10 ** 9 + 7
  
  xor = 0
  xors = []
  for i in range(int(3*10**5)+100):
      xor ^= i
      xors.append(xor)
  
  class Solution():
      def main(self):
          (a, b) = getil()    
          xor = xors[a-1]
          req = b^xor
          return a + (0 if xor==b else 1 if req != a else 2)
  
  
  
  try:
      Sol = Solution()
      Test = geti()
      for _ in range(Test):
          ans = Sol.main()        
          print(ans)
  except Exception as e:
      print(e)
      traceback.print_exc()