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

def input(): return sys.stdin.readline().rstrip("\r\n")

def maps():return [int(i) for i in input().split()]

#lOOKOUT FOR THE EDGE CASES

# print(bin(1000)[2:])
# Python3 implementation of the approach
from operator import xor

# Function to return the XOR of elements
# from the range [1, n]
def findXOR(n):
	mod = n % 4;

	# If n is a multiple of 4
	if (mod == 0):
		return n;

	# If n % 4 gives remainder 1
	elif (mod == 1):
		return 1;

	# If n % 4 gives remainder 2
	elif (mod == 2):
		return n + 1;

	# If n % 4 gives remainder 3
	elif (mod == 3):
		return 0;

# Function to return the XOR of elements
# from the range [l, r]
def findXORFun(l, r):
	return (xor(findXOR(l - 1) , findXOR(r)));

# # Driver code
# l = 4; r = 8;

# print(findXORFun(l, r));

# # This code is contributed by PrinciRaj1992
# print(findXORFun(0,7))

for _ in range(*maps()):
	a , b = maps()
	#a -- mex , b -- xor
	x = findXORFun(0 , a-1)
	n = a 
	if x == b:
		print(n)
	else:
		xi = bin(x)[2:]
		bb = bin(b)[2:]
		if len(bb) < len(xi):
			bb= '0'*(len(xi) - len(bb)) + bb
		if len(xi) < len(bb):
			xi = '0'*(len(bb) - len(xi)) + xi

		A = ['0']*len(xi)
		ok = False
		for i in range(len(xi)):
			if xi[i] == '1' and bb[i] == '1':
				ok = True
				A[i] = '0'
			elif xi[i] == '1' and bb[i] == '0':
				A[i] = '1'
			elif xi[i] == '0' and bb[i] == '1':
				A[i] = '1'
		
		AA = ''.join(A)
		if int(AA,2) != a :
			print(n+1)
		else:
			print(n+2)

