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.util.*;
import java.io.*;

public class C_Mikasa{
    public static void main(String[] args) {
        FastScanner s= new FastScanner();
        StringBuilder res = new StringBuilder();
        int t=s.nextInt();
        int p=0;
        while(p<t){
        int n=s.nextInt();
        int m=s.nextInt();
        if(n>m){
            res.append("0\n");
            p++;
            continue;
        }
        int first[]= new int[32];
        int second[]= new int[32];
        int num1=n;
        int k=0;
        while(num1>0){
            int yo=num1&1;
            first[k]=yo;
            num1=num1>>1;
            k++;

        }
        k=0;
         num1=m;
         while(num1>0){
            int yo=num1&1;
            second[k]=yo;
            num1=num1>>1;
            k++;

        }
        // for(int i=0;i<32;i++){
        //     System.out.print(first[i]+" ");
        // }
        // System.out.println();

        // for(int i=0;i<32;i++){
        //     System.out.print(second[i]+" ");
        // }
        // System.out.println();
        long min=Long.MAX_VALUE;
        long val=0;
        for(int i=31;i>=0;i--){
            if(second[i]==0){
                int num2=1^first[i];
                long pow=0;
                
                  pow=((long)Math.pow(2,i))*num2;
                  min=Math.min(min,(val+pow));
                 
                
               
               
            }
            else{
                long numm=((long)Math.pow(2,i))*(1^first[i]);
                val+=numm;
            }
        }
        res.append(min+"\n");
        p++;
    }
    System.out.println(res);


    }
    static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
 
		public FastScanner(String s) {
			try {
				br = new BufferedReader(new FileReader(s));
			} catch (FileNotFoundException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
		}
 
		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String nextToken() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					// TODO Auto-generated catch block
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(nextToken());
		}
 
		long nextLong() {
			return Long.parseLong(nextToken());
		}
 
		double nextDouble() {
			return Double.parseDouble(nextToken());
		}
	}




}