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
  
  def input(): return sys.stdin.readline().rstrip("\r\n")
  
  def maps():return [int(i) for i in input().split()]
  
  #lOOKOUT FOR THE EDGE CASES
  
  from operator import xor
  
  def findXOR(n):
      mod = n % 4;
  
      # If n is a multiple of 4
      if (mod == 0):
          return n;
  
      # If n % 4 gives remainder 1
      elif (mod == 1):
          return 1;
  
      # If n % 4 gives remainder 2
      elif (mod == 2):
          return n + 1;
  
      # If n % 4 gives remainder 3
      elif (mod == 3):
          return 0;
  
  # Function to return the XOR of elements
  # from the range [l, r]
  def findXORFun(l, r):
      return (xor(findXOR(l - 1) , findXOR(r)));
  
  
  for _ in range(*maps()):
      a , b = maps()
      #a -- mex , b -- xor
      x = findXORFun(0 , a-1)
      if x == b:
          print(a)
      else:
          if x ^ b == a :
              print(a+2)
          else:
              print(a+1)
  
          # xi = bin(x)[2:]
          # bb = bin(b)[2:]
          # if len(bb) < len(xi):
          #     bb= '0'*(len(xi) - len(bb)) + bb
          # if len(xi) < len(bb):
          #     xi = '0'*(len(bb) - len(xi)) + xi
  
          # A = ['0']*len(xi)
          # ok = False
          # for i in range(len(xi)):
          #     if xi[i] == '1' and bb[i] == '1':
          #         ok = True
          #         A[i] = '0'
          #     elif xi[i] == '1' and bb[i] == '0':
          #         A[i] = '1'
          #     elif xi[i] == '0' and bb[i] == '1':
          #         A[i] = '1'
          
          # AA = ''.join(A)
          # if int(AA,2) != a :
          #     print(n+1)
          # else:
          #     print(n+2)
  
  