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
  
  pl=1
  sys.setrecursionlimit(10**5)
  if pl:
  	input=sys.stdin.readline
  else:	
  	sys.stdin=open('input.txt', 'r')
  	sys.stdout=open('outpt.txt','w')
  
  def li():
  	return [int(xxx) for xxx in input().split()]
  def fi():
  	return int(input())
  def si():
  	return list(input().rstrip())	
  def mi():
  	return 	map(int,input().split())	
  def ff():
  	sys.stdout.flush()
  def google(tc,*ans):
  	print("Case #"+str(tc)+":",*ans)	
  def bits(i,n):
  	p=bin(i)[2:]
  	return (n-len(p))*"0"+p			
  t=fi()
  f=t	
  k=3*10**5+2
  pre=[0]*k
  for i in range(k):
  	pre[i]=pre[i-1]^i
  while t:
  	t-=1
  	mex,xor=mi()
  	x=pre[mex-1]
  	u=x^xor
  	if u==0:
  		ans=mex
  	elif u==mex:
  		ans=mex+2
  	else:
  		ans=mex+1	
  	print(ans)	