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.
import java.io.*;
import java.util.*;
public class Main {

    public static long helper(String s){
        long score=0;
        int count=1;
        for(int i=1;i<s.length();i++){
            if(s.charAt(i)==s.charAt(i-1)){
                count+=1;
            }
            else{
                score+=((long)count*count);
                count=1;
            }
        }


        return score+=(count*count);
    }


    public static void main(String[] args) {
        FastScanner fs=new FastScanner();
        PrintWriter out=new PrintWriter(System.out);
        int t=fs.nextInt();
        while(t-->0){
            int n=fs.nextInt();
            int m= fs.nextInt()+1;

            int fa=0;

            int i=Integer.highestOneBit(m);
            i = (int)(Math.log(i) / Math.log(2));
            while(i>=0){
                if((n^fa)>m)
                    break;
                if((m&(1<<i))!=0){
                    if((n&(1<<i))!=0){
                        i-=1;
                        continue;
                    }
                    else{
                        fa+=(1<<i);
                        i-=1;
                        continue;
                    }
                }
                else{
                    i-=1;
                }
            }
            out.println(fa);


        }
        out.close();

//        int f[]=fs.readArray((1<<n));
//
//        int dp[][]=new int[(1<<n)][n+1];
//
//        for(int t=0;t<(1<<n);t++){
//            dp[t][0]=f[t];
//        }
//        for(int t=0;t<(1<<n);t++){
//            for(int i=1;i<n+1;i++){
//                if((t&(1<<(i-1)))!=0){
//                    dp[t][i]=dp[t][i-1]+dp[t^(1<<(i-1))][i];
//                }
//                else{
//                    dp[t][i]=dp[t][i-1];
//                }
//            }
//        }
//        for(int i=0;i<(1<<n);i++){
//            out.print(dp[i][n]+" ");
//        }

        out.println();
        out.close();




       






    }
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next() {
            while (!st.hasMoreElements())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }
}
