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 java.io.*;
import java.util.*;

public class Demo{

    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer st = null;
    static int mod = 1000_000_007;

    public static void main(String[] args) throws IOException{
        int testcase = Integer.parseInt(br.readLine());
        int[] xors = new int[300001];
        int temp = 0;
        for(int i=0;i<300001;i++){
            temp^=i;
            xors[i] = temp;
        }
        while(testcase-->0){
            st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken());
            int m = Integer.parseInt(st.nextToken());
            int t = xors[n-1];
            int ans = n;

            if(t!=m){
                if(t==0 && n==m)ans+=2;
                else if((t^m)==n){
                    ans+=3;
                }else{
                    ans++;
                }
            }
            
            System.out.println(ans);
            
        }

    }
}
/*

*/