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.
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)	