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.
///In the name of ALLAH at the beginning and at the end.
#include<bits/stdc++.h>
using namespace std;

typedef long long   ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int,int> pii;
typedef pair<double, double> pdd;
typedef pair<ll, ll> pll;
typedef vector<pii> vii;
typedef vector<pll> vll;
typedef double dl;

#define PB push_back
#define F first
#define S second
#define MP make_pair
#define endl '\n'
#define all(a) (a).begin(),(a).end()
#define sz(x) (int)x.size()
#define mid(l,r) ((r+l)/2)
#define left(node) (node*2)
#define right(node) (node*2+1)
#define mx_int_prime 999999937

const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const ll infLL = 9000000000000000000;
#define MOD 1000000007

#define mem(a,b) memset(a, b, sizeof(a) )
#define gcd(a,b) __gcd(a,b)
#define sqr(a) ((a) * (a))

#define optimize() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define fraction() cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);
#define file() freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;

typedef tree< ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update > ordered_set;
/// change ll to any data type
/// less_equal for multiset increasing order
/// less for set increasing order
/// greater_equal for multiset decreasing order
/// greater for set decreasing order

/// cout<<*X.find_by_order(1)<<endl; // iterator to the k-th  element
/// cout<<X.order_of_key(-5)<<endl;  // number of items in a set that are strictly smaller than our item

typedef vector<int>::iterator vit;
typedef set<int>::iterator sit;


int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
//int dx[] = {+1, 0, -1, 0, +1, +1, -1, -1};
//int dy[] = {0, +1, 0, -1, +1, -1, +1, -1};



template < typename F, typename S >
ostream& operator << ( ostream& os, const pair< F, S > & p )
{
    return os << "(" << p.first << ", " << p.second << ")";
}

template < typename T >
ostream &operator << ( ostream & os, const vector< T > &v )
{
    os << "{";
    for(auto it = v.begin(); it != v.end(); ++it)
    {
        if( it != v.begin() ) os << ", ";
        os << *it;
    }
    return os << "}";
}

template < typename T >
ostream &operator << ( ostream & os, const set< T > &v )
{
    os << "[";
    for(auto it = v.begin(); it != v.end(); ++it)
    {
        if( it != v.begin() ) os << ", ";
        os << *it;
    }
    return os << "]";
}

template < typename T >
ostream &operator << ( ostream & os, const multiset< T > &v )
{
    os << "[";
    for(auto it = v.begin(); it != v.end(); ++it)
    {
        if( it != v.begin() ) os << ", ";
        os << *it;
    }
    return os << "]";
}

template < typename F, typename S >
ostream &operator << ( ostream & os, const map< F, S > &v )
{
    os << "[";
    for(auto it = v.begin(); it != v.end(); ++it)
    {
        if( it != v.begin() ) os << ", ";
        os << it -> first << " = " << it -> second ;
    }
    return os << "]";
}

#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

void faltu ()
{
    cerr << endl;
}

template <typename T>
void faltu( T a[], int n )
{
    for(int i = 0; i < n; ++i) cerr << a[i] << ' ';
    cerr << endl;
}

template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest)
{
    cerr << arg << ' ';
    faltu(rest...);
}

/**********************Code starts here*********************************/
int main ()
{
    optimize();
    int t  ; cin >> t ;
    while ( t-- )
    {
        int a , b ;
        cin >> a >> b ;
        if ( a == 1 and b == 1 )
        {
            cout << 3 << endl ; continue ;
        }
        if ( b == 0 )
        {
            cout << a + 1 << endl ; continue ;
        }
        if ( b ==  1 and a == 2 )
        {
            cout << 2 << endl ;
            continue ;
        }
        else
            cout << a + 1 << endl ;
    }
    return 0;
}
