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 CF735C {
  
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  		Scanner s= new Scanner(System.in);
  		int t=s.nextInt();
  		while(t-->0) {
  			int n=s.nextInt();
  			int m=s.nextInt();
  			solve(n,m);
  		}
  
  	}
  	static void solve(int n,int m) {
  //		System.out.println(n+" "+m);
  		int bit=32;
  		int ans=0;
  		//n^k>=(m+1)
  		for(int i = bit-1;i>=0;i--) {
  			//n=1 and m+1 =1
  			int a=n&(1<<i);
  			int b=(1<<i)&(m+1);
  //			System.out.println("i:"+i+" a:"+a+" b:"+b);
  			if(a>=1 && b>=1){
  				
  			}
  			//n=0 and m+1 =1
  			else if(a==0 && b>=1){
  //				System.out.println("here");
  				ans+=(1<<i);
  			}
  			//n=1 and m+1 =0
  			else if(a>=1 && b==0){
  				break;
  			}
  			//n=0 and m+1 =0
  			else{
  				
  			}
  		}
  		System.out.println(ans);
  	}
  
  }
  