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, os
  from io import BytesIO, IOBase
  from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
  from collections import defaultdict as dd, deque
  from heapq import merge, heapify, heappop, heappush, nsmallest
  from bisect import bisect_left as bl, bisect_right as br, bisect
   
  # region fastio
  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")
   
  stdin, stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) 
   
  mod = pow(10, 9) + 7
  mod2 = 998244353
          
  def inp(): return stdin.readline().strip()
  def iinp(): return int(inp())
  def out(var, end="\n"): stdout.write(str(var)+"\n")
  def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
  def lmp(): return list(mp())
  def mp(): return map(int, inp().split())
  def l1d(n, val=0): return [val for i in range(n)]
  def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
  def ceil(a, b): return (a+b-1)//b
  S1 = 'abcdefghijklmnopqrstuvwxyz'
  S2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
  def isprime(x):
      if x<=1: return False
      if x in (2, 3): return True
      if x%2 == 0: return False
      for i in range(3, int(sqrt(x))+1, 2):
          if x%i == 0: return False
      return True
   
  N = 300005 
  xor = l1d(N)
  for i in range(1, N):
      xor[i] = xor[i-1]^i
  
  for _ in range(iinp()):
      a, b = mp()
      ans = a
      if xor[a-1] == b:
          print(ans)
      elif xor[a] == b:
          print(ans+2)
      else:
          print(ans+1)