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 MainC {
      static class InputReader {
          public BufferedReader reader;
          public StringTokenizer tokenizer;
          public InputReader(InputStream stream) {
              reader = new BufferedReader(new InputStreamReader(stream), 32768);
              tokenizer = null;
          }
          public String next() {
              while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                  try {
                      tokenizer = new StringTokenizer(reader.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return tokenizer.nextToken();
          }
          public int nextInt() {
              return Integer.parseInt(next());
          }
          public long nextLong() {
              return Long.parseLong(next());
          }
          public boolean hasNext() {
              try {
                  String string = reader.readLine();
                  if (string == null) {
                      return false;
                  }
                  tokenizer = new StringTokenizer(string);
                  return tokenizer.hasMoreTokens();
              } catch (IOException e) {
                  return false;
              }
          }
      }
      static InputReader in = new InputReader(System.in);
      static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
  
      static int[] pow = new int[31];
      static void ini () {
          pow[0] = 1;
          for (int i = 1; i < 31; i++) {
              pow[i] = 2 * pow[i-1];
          }
      }
  
      public static void main(String[] args) {
          int t = in.nextInt();
          ini();
          while (t -- > 0) {
              solve();
          }
          out.close();
      }
      static void solve() {
          int n = in.nextInt();
          int m = in.nextInt();
         
          int res = 0;
          m ++;
          for (int i = 30; i >= 0; i--) {
              int a = (n >> i) & 1;
              int b = (m >> i) & 1;
              if (a > b) {
                  out.println(res);
                  return;
              }
              if (a < b) {
                  res |= (1 << i);
              }
          }
          out.println(res);
          
      }
  }