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.
from typing import Counter
import sys
from collections import defaultdict as dd
from math import *
 
def vinp():
    return map(int,input().split())
def linp():
    return list(map(int,input().split()))
def sinp():
    return input()
def inp():
    return int(input())
def mod(f):
    return f % 1000000007
def pr(*x):
    print(*x)
def finp():
    f=open("input.txt","r")
    f=f.read().split("\n")
    return f
def finp():
    f=open("input.txt","r")
    f=f.read().split("\n")
    return f
def fout():
    return open("output.txt","w")
def fpr(f,x):
    f.write(x+"\n")
def csort(c):
    sorted(c.items(), key=lambda pair: pair[1], reverse=True)
def indc(l,n):
    c={}
    for i in range(n):
        c[l[i]]=c.get(l[i],[])+[i+1]
    return c
def XOR(n) :
    if n % 4 == 0 :
        return n
    if n % 4 == 1 :
        return 1
    if n % 4 == 2 :
        return n + 1
    return 0
if __name__ =="__main__":
    cou=inp()
    for i in range(cou):
        a,b = vinp()
        z = XOR(a-1)
        z = z^b
        if z!=0:
            if z==a:
                a+=2
            else:
                a+=1
        pr(a)