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 os
  import sys
  from io import BytesIO, IOBase
  #import sys
  #input = sys.stdin.buffer.readline
  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, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  #######################################
  from itertools import permutations
  from collections import *
  from collections import deque
  from operator import itemgetter , attrgetter
  from decimal import *
  import bisect
  import math
  import heapq as hq
  #import sympy
  MOD=10**9 +7
  
  
  def is_prime(n):
    if n == 2 or n == 3: return True
    if n < 2 or n%2 == 0: return False
    if n < 9: return True
    if n%3 == 0: return False
    r = int(n**0.5)
    # since all primes > 3 are of the form 6n ± 1
    # start with f=5 (which is prime)
    # and test f, f+2 for being prime
    # then loop by 6.
    f = 5
    while f <= r:
  
      if n % f == 0: return False
      if n % (f+2) == 0: return False
      f += 6
    return True
  
  def pow(a,b,m):
      ans=1
      while b:
          if b&1:
              ans=(ans*a)%m
          b//=2
          a=(a*a)%m
      return ans
  #vis=[]
  #graph=[]
  
  def ispalindrome(s):
      if s[:]==s[::-1]:
          return 1
      return 0
  dp=[]
  limit=[]
  v=[]
  def dpdfs(u,t=-1):
      dp[0][u]=0
      dp[1][u]=0
      for i in v[u]:
          if i==t:
  
              continue
          if dp[1][i]==-1:
              dpdfs(i,u)
          dp[0][u]+=max(abs(limit[0][u]-limit[1][i])+dp[1][i],abs(limit[0][u]-limit[0][i])+dp[0][i])
          dp[1][u] += max(abs(limit[1][u] - limit[1][i]) + dp[1][i], abs(limit[1][u] - limit[0][i]) + dp[0][i])
  vis=[]
  f=0
  
  from queue import PriorityQueue
  
  def z_algorithm(s):
      res = [0] * len(s)
      res[0] = len(s)
      i, j = 1, 0
      while i < len(s):
          while i + j < len(s) and s[j] == s[i + j]:
              j += 1
          res[i] = j
          if j == 0:
              i += 1
              continue
          k = 1
          while i + k < len(s) and k + res[k] < j:
              res[i + k] = res[k]
              k += 1
          i, j = i + k, j - k
      return res
  def gcd(a, b):
      if a == 0:
          return b
      return gcd(b % a, a)
  
  
  # Function to return LCM of two numbers
  def lcm(a, b):
      return (a / gcd(a, b)) * b
  
  
  def gcdExtended(a, b):
      # Base Case
      if a == 0:
          return b, 0, 1
  
      gcd, x1, y1 = gcdExtended(b % a, a)
  
      # Update x and y using results of recursive
      # call
      x = y1 - (b // a) * x1
      y = x1
  
      return gcd, x, y
  import re
  for _ in range(int(input())):
      me,x=map(int,input().split())
      ans=me 
      
     
          
      if me%4==0:
          t=0
      elif me%4==1:
          t=me-1 
      elif me%4==2:
          t=1 
      else:
          t=me 
      
      if t==x:
          print(ans) 
      elif t^x==me:
          print(ans+2)
      else:
          print(ans+1)
          
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  