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
  
  # sys.setrecursionlimit(200005)
  int1 = lambda x: int(x)-1
  p2D = lambda x: print(*x, sep="\n")
  def II(): return int(sys.stdin.readline())
  def LI(): return list(map(int, sys.stdin.readline().split()))
  def LLI(rows_number): return [LI() for _ in range(rows_number)]
  def LI1(): return list(map(int1, sys.stdin.readline().split()))
  def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
  def SI(): return sys.stdin.readline().rstrip()
  inf = 10**16
  md = 10**9+7
  # md = 998244353
  
  def solve():
      a, b = LI()
      if a == 1:
          if b == 0: ans = 1
          elif b == 1: ans = 3
          else: ans = 2
      else:
          if b == cs[a-1]: ans = a
          elif cs[a-1] ^ b == a: ans = a+2
          else: ans = a+1
      print(ans)
  
  cs = [0]
  for i in range(1, 300005): cs.append(cs[-1] ^ i)
  
  for testcase in range(II()):
      solve()
  