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
  import math
  from bisect import bisect_left
  import heapq
  from collections import deque
  from itertools import product, permutations
  import random
  def II():
  	return int(sys.stdin.readline())
   
  def LI():
  	return list(map(int, sys.stdin.readline().split()))
   
  def MI():
  	return map(int, sys.stdin.readline().split())
   
  def SI():
  	return sys.stdin.readline().strip()
   
  def C(n, k, mod):
      return (FACT(n,mod) * pow((FACT(k,mod)*FACT(n-k,mod))%mod,mod-2, mod))%mod
   
  def lcm(a,b):
      return abs(a*b) // math.gcd(a, b)
  
  x = []
  xor = 0
  for i in range(10**6):
  	xor^=i
  	x.append(xor)
   
  for _ in range(II()):
  	a,b = MI()
  	xor = x[a-1]
  	if xor == b:
  		print(a)
  	elif xor^b == a:
  		print(a+2)
  	else:
  		print(a+1)