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 sys
  import os
  from io import BytesIO
  
  if sys.version_info[0] < 3:
      sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
  
  f = sys.stdin
  if os.environ.get('USER') == "loic":
      f = open("data.in")
  
  line = lambda: f.readline().strip('\r\n').split()
  
  def write(w):
      sys.stdout.write(w)
      sys.stdout.write("\n")
  
  def sumxor(n):
      if n%4==0:
          return n
      if n%4==1:
          return 1
      if n%4==2:
          return n+1
      return 0
  
  def solve():
      s = sumxor(MEX-1)
      d = s^XOR
      
      res = 0
      if d == 0:
          res = MEX
      elif d < MEX:
          res = MEX + 1
      elif d > MEX:
          res = MEX + 1
      else:
          res = MEX + 2
      
      return str(res)
  
  T = int(line()[0])
  for test in range(1,T+1):
      MEX,XOR = map(int,line())
      write(solve())
      
  f.close()