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

public class Solution {

    public static void main(String[] args) throws IOException {

        Scanner in = new Scanner(System.in);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuffer out = new StringBuffer();

        int T = in.nextInt();

        int max = 300_000;
        int xor[] = new int[max+1];
        for(int i=1; i<=max; i++) {
            xor[i]^=(xor[i-1]^i);
        }

        OUTER:
        while (T-- > 0) {

            int a = in.nextInt(),
                b = in.nextInt();

            int ans = -1;
            if(xor[a-1]==b) {
                ans = a;
            } else if((xor[a-1]^b)!=a) {
                ans = a+1;
            } else {
                ans = a+2;
            }
            out.append(ans+"\n");
        }

        System.out.print(out);
    }

    private static void sort(List<Integer> list) {
        list.stream().sorted();
    }

    private static long gcd(long a, long b) {
        if (a==0)
            return b;
        return gcd(b%a, a);
    }

    private static int toInt(String s) {
        return Integer.parseInt(s);
    }

    private static long toLong(String s) {
        return Long.parseLong(s);
    }
}