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
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 b == a == 1:
		print(a+2)
	else:
		print(a+1)