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 GFG {
      
      public static class Pair{
          long x;
          int y;
          Pair(long x,int y){
              this.x = x;
              this.y = y;
          }
      }
      
  	public static void main (String[] args) {
  	    Scanner sc = new Scanner(System.in);
  	    int t = sc.nextInt();
  	     while(t-- > 0){
  	         long n = sc.nextLong();
  	         long m = sc.nextLong();
  	         long prev = 0,ans = Long.MAX_VALUE;
  	         for(long i=32;i>=0;i--)
  	         {
  	             long bit1 = (1L<<i)&n;
  	             long bit2 = (1L<<i)&m;
  	             if(bit2>0){
  	                 if(bit1==0)
  	                 prev |= (1L<<i);
  	             }
  	             else{
  	                 if(bit1==0)
  	                 ans = Math.min(ans,prev|(1L<<i));
  	                 else{
  	                     ans = Math.min(ans,prev);
  	                     prev|=bit1;
  	                 }
  	             }
  	         }
  	         System.out.println(ans);
  	     }    
  	}
  }
  