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.util.Scanner;
  
  public class Main {
  
      public static void main(String[] agrs) {
          int bit = 31;
          Scanner sc = new Scanner(System.in);
          int testCase = sc.nextInt();
          for(int i=0; i<testCase; i++) {
             int n = sc.nextInt();
             int m = sc.nextInt();
              int ans = 0;
             for(int j=bit-1; j>=0; j--) {
                 //n->1 & m+1->1
                 if((1<<j & n) != 0 &&  ((1<<j)&(m+1)) != 0) {
                 }
                 //n->0 & m+1->1
                 else if(((1<<j)&n) == 0 && ((1<<j)&(m+1)) > 0) {
                     ans =ans | (1<<j);
                 }
                 //n->1 & m+1->0
                 else if (((1<<j) & n) > 0 && ((1<<j)&(m+1)) == 0) {
                     break;
                 }
                 //n->0 & m+1->0
                 else {
  
                 }
             }
             System.out.println(ans);
          }
      }
  }