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.ArrayList;
import java.util.Scanner;

public class Solution{
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        ArrayList<Integer> fast = new ArrayList<Integer>();
        fast.add(0);
        int fastLen = 1;
        for(int i = 0; i < n; i++){
            int a = in.nextInt();
            int b = in.nextInt();
            while(fastLen < a) {
            	fast.add(fast.get(fastLen - 1) ^ fastLen++);
            }
            int xor = fast.get(a - 1);
            int len = a;
            if(xor != b) {
                if((b ^ xor) == a) {
                	len+=2;
                }else{
                	len++;
                }
            }
            System.out.println(len);
        }
        in.close();
    }
}