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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.math.BigInteger;
  import java.util.*;
  import javafx.util.Pair;
  
  public class Main
  {
    
      public static void main(String[] args)
      {
          FastScanner input = new FastScanner();
          int tc = input.nextInt();
          while (tc-- > 0) {
            
         int bit = 31;
         int n = input.nextInt();
         int m = input.nextInt()+1;
         int ans =0;
          for (int i = bit-1; i >=0; i--) {
  //            System.out.println("/");
  //            System.out.println("i->"+i);
  //            System.out.println((n>>i)+" "+(m>>i));
             int bit1 = ((n>>i)&1);
             int bit2 = ((m>>i)&1);
  //            System.out.println(bit1+" "+bit2);
              if(bit1==bit2)
              {
                  continue;
              }
              else if(bit1==1)
              {
                  break;
              }
              else
              {
                  ans+=1<<i;
              }
          }
          System.out.println(ans);   
          }
         
      }
  
      static class FastScanner
      {
  
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
  
          String next()
          {
              while (!st.hasMoreTokens()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                  }
              }
              return st.nextToken();
          }
  
          int nextInt()
          {
              return Integer.parseInt(next());
          }
  
          long nextLong()
          {
  
              return Long.parseLong(next());
          }
  
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
  
          String nextLine() throws IOException
          {
              return br.readLine();
          }
      }
  
  }
  