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.StringTokenizer;

public class MEXorMixup {
    static int XOR(int n) {
        if (n % 4 == 0)
            return n;
        if (n % 4 == 1)
            return 1;
        if (n % 4 == 2)
            return n + 1;
        return 0;
    }

    public static void main(String[] args) throws FileNotFoundException {
        FastReader in = new FastReader();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

        int t = in.nextInt();
        while (t-- > 0) {
            long mex = in.nextInt();
            long xor = in.nextInt();
            long res = mex;

            int x = XOR((int) (mex - 1));

            if (x == xor)
                ;
            else if ((mex ^ x) == xor)
                res += 2;
            else
                res++;

            out.println(res);
        }

        out.close();
    }

    private static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        FastReader() throws FileNotFoundException {
            br = new BufferedReader(new
                    InputStreamReader(System.getProperty("ONLINE_JUDGE") == null ? new FileInputStream("input.txt") : System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}