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: 
  def ii():  return int(input())
  def si():  return input()
  def mi():  return map(int,input().strip().split(" "))
  def msi(): return map(str,input().strip().split(" "))
  def li():  return list(mi())
  
  
  
  # for _ in range(ii()):
  #     n = ii()
  #     s = input()
  #     ans = ""
  #     for i in s:
  #         if(i=='U'):
  #             ans += "D"
  #         elif(i=='D'):
  #             ans += "U"
  #         else:
  #             ans += i
  #     print(ans)
  
  for _ in range(ii()):
      a,b = mi()
      mex = a-1
      x = 0
      if(mex):
          if(mex%4 == 1):
              x = 1
          elif(mex%4 == 2):
              x = (mex//4 + 1)*4 - 1
          elif(mex%4 == 3):
              x = 0
          else:
              x = mex
  
      ans = 0
      if(x==b):
          ans = a
      elif(x^b == a):
          ans = a + 2
      else:
          ans = a+1
      print(ans)