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
try:
	sys.stdin = open('inputf.in', 'r')
	sys.stdout=open('outputf.in','w')	
except:
	pass
#------------------------------------------------
#importing libraries
import math
import sys
from math import sqrt
from collections import defaultdict
from collections import Counter
import string
from collections import deque
from functools import cmp_to_key
import bisect
from collections import OrderedDict
import heapq as hq
from queue import PriorityQueue
#----------------------------------------------------------------------
#helper Module
# sys.setrecursionlimit(1000000)
MOD=(10**9)+7
#----------------------------------------------------------------------
#input snippets
def ii():return int(sys.stdin.readline().strip())
def si():return sys.stdin.readline().strip()
def li():return list(map(int,sys.stdin.readline().strip().split()))
def lsi():return list(map(str,sys.stdin.readline().strip().split()))
def mi():return map(int,sys.stdin.readline().strip().split())
def dparr(m,n): return [[0]*n for i in range(m)]
#----------------------------------------------------------------------
def computeXOR(n) :
 
	# Modulus operator are expensive
	# on most of the computers. n & 3
	# will be equivalent to n % 4.
 
	# if n is multiple of 4
	if n % 4 == 0 :
		return n
 
	# If n % 4 gives remainder 1
	if n % 4 == 1 :
		return 1
 
	# If n%4 gives remainder 2
	if n % 4 == 2 :
		return n + 1
 
	# If n%4 gives remainder 3
	return 0
t=ii()
for _ in range(t):
	a,b=mi()
	ob=b
	xv=computeXOR(a-1)	
	oxv=xv
	# print(xv)
	dq=deque()
	while(xv>0 and b>0):
		if(xv&1 != b&1):
			if(xv&1==0 and b%1==1):
				dq.appendleft("0")
			if(xv&1==1 and b&1==0):
				dq.appendleft("1")
		else:
			dq.appendleft(str(b&1))
		xv=xv>>1
		b=b>>1
	
	while(xv>0):
		if(xv&1==0):
			dq.appendleft("1")
		else:
			dq.appendleft("0")
		xv=xv>>1
	while(b>0):
		if(b&1==0):
			dq.appendleft("0")
		else:
			dq.appendleft("1")
		b=b>>1
	# print(dq)
	
	# print(oxv,x)
	# print(ob)
	if(oxv==ob):
		print(a)
	else:
		x=int("".join(list(dq)),2)
		if(x==a):
			print(a+2)
		else:
			print(a+1)




