description: 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.
solution: import java.io.*;
  import java.util.*;
  
  public class C {
  	
  	public static void main(String[] args) {
  		
  		FastReader f = new FastReader();
  		StringBuffer sb=new StringBuffer();
  		int test=f.nextInt();
  		out:
  		while(test-->0)
  		{	
  			int n=f.nextInt();
  			int m=f.nextInt();
  			
  			if(m<n)
  			{
  				sb.append("0\n");
  				continue;
  			}
  			int nD[]=decToBinary(n);
  							
  			int curr=0,extra=0;;
  			while(true)
  			{	
  				curr=0;
  				for(int i=0;i<32;i++)
  					if(nD[i]==1)
  						curr+=(int)Math.pow(2,i);
  				if(curr>m)
  					break;
  				
  				extra=0;
  				for(int i=0;i<32;i++)
  				{
  					if(nD[i]==0)
  					{
  						extra+=(int)Math.pow(2,i);
  						if((curr)+extra>m)
  						{
  							nD[i]=1;
  							break;
  						}
  					}		
  				}
  			}
  			sb.append((n^curr)+"\n");
  		}
  		System.out.println(sb);
  	}
  	
  	public static int[] decToBinary(int n)
      {	
  		int a[]=new int[32];
          for (int i=31;i>=0;i--) {
              int k=n>>i;
              if((k&1)>0)
              	a[i]=1;
              else
              	a[i]=0;
          }
          return a;
      }
  	
  	static class FastReader 
  	{ 
  	    BufferedReader br; 
  	    StringTokenizer st; 
   
  	    public FastReader() {
  	    	br = new BufferedReader(new
  	                 InputStreamReader(System.in)); 
  	    }
  	    String next() { 
  	        while (st == null || !st.hasMoreElements()) { 
  	            try{ 
  	                st = new StringTokenizer(br.readLine()); 
  	            } 
  	            catch (IOException  e) { 
  	                e.printStackTrace(); 
  	            } 
  	        } 
  	        return st.nextToken(); 
  	    } 
  	    int nextInt() { 
  	        return Integer.parseInt(next()); 
  	    } 
  	    long nextLong() { 
  	        return Long.parseLong(next()); 
  	    } 
  	    double nextDouble() { 
  	        return Double.parseDouble(next()); 
  	    } 
  	    String nextLine() { 
  	        String str = ""; 
  	        try{ 
  	            str = br.readLine(); 
  	        } 
  	        catch (IOException e) { 
  	            e.printStackTrace(); 
  	        } 
  	        return str; 
  	    } 
  	} 
  }