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 B {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringBuilder bw = new StringBuilder();
		int[] P = new int[300001];
		for (int i = 1; i <= 300000; i++) P[i] = i ^ P[i - 1];
		int TC = Integer.parseInt(br.readLine());
		while (TC -- > 0) {
			StringTokenizer st = new StringTokenizer(br.readLine(), " ");
			int a = Integer.parseInt(st.nextToken());
			int b = Integer.parseInt(st.nextToken());
			int ret = a;
			if (P[a - 1] != b) {
				ret++;
				if ((P[a - 1] ^ a) == b) ret++;
			}
			bw.append(ret).append("\n");
		}
		System.out.print(bw);
	}

}