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: // package codeforce.cf735;
  
  import java.io.PrintWriter;
  import java.util.Scanner;
  
  public class C {
      //    MUST SEE BEFORE SUBMISSION
  //    check whether int part would overflow or not, especially when it is a * b!!!!
  
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          PrintWriter pw = new PrintWriter(System.out);
          int t = sc.nextInt();
  //        int t = 1;
          for (int i = 0; i < t; i++) {
              solve(sc, pw);
          }
          pw.close();
      }
  
      static void solve(Scanner in, PrintWriter out){
          int n = in.nextInt(), m = in.nextInt();
          boolean isP = true;
          int v = 0;
          m++;
          for (int i = 31; i >= 0 ; i--) {
              int vn = 0, vm = 0;
              if ((n | (1 << i)) == n) vn = 1;
              if ((m | (1 << i)) == m) vm = 1;
              if (isP){
                  if (vm == 1){
                      v += (1 << i);
                  }else if (vn == 0){
  
                  }else{
                     v += (1 << i);
                     isP = false;
                  }
              }else{
                  if (vn == 1){
                      v += (1 << i);
                  }
              }
          }
          out.println((v ^ n));
      }
  }
  