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.
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import static java.lang.Math.*;
import java.util.*;
import java.io.*;
import java.math.*;

public class x1554C
{
    public static void main(String hi[]) throws Exception
    {
        BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(infile.readLine());
        int T = Integer.parseInt(st.nextToken());
        StringBuilder sb = new StringBuilder();
        while(T-->0)
        {
            st = new StringTokenizer(infile.readLine());
            int N = Integer.parseInt(st.nextToken());
            int M = Integer.parseInt(st.nextToken());
            if(N > M)
                sb.append("0\n");
            else if((N&M) == N)
            {
                int res = N^M;
                for(int b=0; b < 30; b++)
                {
                    if((res&(1<<b)) == 0)
                    {
                        res |= (1<<b);
                        break;
                    }
                    res ^= 1<<b;
                }
                sb.append(res+"\n");
            }
            else
            {
                int res = 0;
                for(int b=29; b >= 0; b--)
                {
                    if((N&(1<<b)) > 0 && (M&(1<<b)) == 0)
                        break;
                    if((N&(1<<b)) > 0 && (M&(1<<b)) > 0)
                        continue;
                    res |= (M&(1<<b));
                }
                sb.append(res+"\n");
            }
            /*
            boolean[] lol = new boolean[1000];
            for(int a=0; a <= M; a++)
                lol[N^a] = true;
            for(int a=0; a < 1000; a++)
                if(!lol[a])
                {
                    System.out.println("Supposed to be: "+a);
                    break;
                }
                
             */
        }
        System.out.print(sb);
    }
    public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
    {
        int[] arr = new int[N];
        st = new StringTokenizer(infile.readLine());
        for(int i=0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
        return arr;
    }
}
/*
0 1 2 3 4 5
2 3 0 1 6 7

3 2 1 0 7 6
MEX is at most M+1

00011110001001000000
10011111101111110001

10000001100110110001

 */