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.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  
  public class C{
      public static void main(String[] args){
          FS sc = new FS();
          PrintWriter pw = new PrintWriter(System.out);
          int t = sc.nextInt();
          while(t-->0){
              int n = sc.nextInt();
              int m = sc.nextInt();
              m = m+1;
              int k = 0;
              for(int i=31; i>=0; i--){
                  if((n&(1<<i))==0 && (m&(1<<i))!=0) k = k | (1<<i);
                  else if((n&(1<<i))!=0 && (m&(1<<i))==0) break;
              }
              System.out.println(k);
          }
          pw.flush();
          pw.close();
      }
  
      static class FS{
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
      
          String next(){
              while(!st.hasMoreElements()){
                  try{
                      st = new StringTokenizer(br.readLine());
                  } catch(Exception ignored){
                  }
              }
              return st.nextToken();
          }
          
          int[] nextArray(int n){
              int[] a = new int[n];
              for(int i = 0; i < n; i++){
                  a[i] = nextInt();
              }
              return a;
          }
          
          long[] nextLongArray(int n){
              long[] a = new long[n];
              for(int i = 0; i < n; i++){
                  a[i] = nextLong();
              }
              return a;
          }
          
          int nextInt(){
              return Integer.parseInt(next());
          }
          
          long nextLong(){
              return Long.parseLong(next());
          }
          
          double nextDouble(){
              return Double.parseDouble(next());
          }
      }
  }