You are given two integers n and m. Find the \operatorname{MEX} of the sequence n ⊕ 0, n ⊕ 1, …, n ⊕ m. Here, ⊕ is the [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

\operatorname{MEX} of the sequence of non-negative integers is the smallest non-negative integer that doesn't appear in this sequence. For example, \operatorname{MEX}(0, 1, 2, 4) = 3, and \operatorname{MEX}(1, 2021) = 0. 

Input

The first line contains a single integer t (1 ≤ t ≤ 30 000) — the number of test cases.

The first and only line of each test case contains two integers n and m (0 ≤ n, m ≤ 10^9).

Output

For each test case, print a single integer — the answer to the problem.

Example

Input


5
3 5
4 6
3 2
69 696
123456 654321


Output


4
3
0
640
530866

Note

In the first test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, 3 ⊕ 3, 3 ⊕ 4, 3 ⊕ 5, or 3, 2, 1, 0, 7, 6. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 4.

In the second test case, the sequence is 4 ⊕ 0, 4 ⊕ 1, 4 ⊕ 2, 4 ⊕ 3, 4 ⊕ 4, 4 ⊕ 5, 4 ⊕ 6, or 4, 5, 6, 7, 0, 1, 2. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 3.

In the third test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, or 3, 2, 1. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 0.
// package codeforce.cf735;

import java.io.PrintWriter;
import java.util.Scanner;

public class C {
    //    MUST SEE BEFORE SUBMISSION
//    check whether int part would overflow or not, especially when it is a * b!!!!

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
//        int t = 1;
        for (int i = 0; i < t; i++) {
            solve(sc, pw);
        }
        pw.close();
    }

    static void solve(Scanner in, PrintWriter out){
        int n = in.nextInt(), m = in.nextInt();
        boolean isP = true;
        int v = 0;

        for (int i = 31; i >= 0 ; i--) {
            int vn = 0, vm = 0;
            if ((n | (1 << i)) == n) vn = 1;
            if ((m | (1 << i)) == m) vm = 1;
            if (isP){
                if (vm == 1){
                    v += (1 << i);
                }else if (vn == 0){

                }else{
                   v += (1 << i);
                   isP = false;
                }
            }else{
                if (vn == 1){
                    v += (1 << i);
                }
            }
        }
        if (v == m){
            for(int i = 0; i < 32; i++){
                if ((m | (1 << i)) != m) {
                    v += (1 << i);
                    break;
                }
            }
            out.println(((m + 1) ^ n));
        }else{
            out.println((v ^ n));
        }
    }
}
