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.*;
      public class Solution
      {
      	public static void main(String args[])
      	{
      		Scanner sc=new Scanner(System.in); 
      		int test= sc.nextInt();
      		for(int xix=1;xix<=test;xix++)
      		{
      			int n=sc.nextInt(); int m=sc.nextInt();
      			if(n>m)
      			{
      				System.out.println(0);
      			}
      			else
      			{
      				String q=Integer.toBinaryString(((n^m)&n));
      				String s=Integer.toBinaryString(m);
      				String ans=s.substring(0,s.length()-q.length())+q;
      				if(q.equals("0"))
      				{
      					ans=s;
      				}
      				int r=Integer.parseInt(ans,2);
      				r|=n;
      				if(r==m)
      				{
      					r++;
      					r|=n;
      					if((r^n)>r)
      					{
      						System.out.println(r);
      					}
      					else
      					{
      						System.out.println(r^n);
      					}
      				}
      				else
      				{
      					System.out.println(r^n);
      				}
      			}
      		}
      		sc.close();
      	}
      }