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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  
  public class C_Rnd735_Mikasa 
  {
  	public static void main(String[] args) throws NumberFormatException, IOException
  	{
  		BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
  		PrintWriter out = new PrintWriter(System.out);
  
  		String[] in;
  		int tt = Integer.parseInt(scan.readLine());
  		
  		for(int t = 0; t < tt; t++)
  		{
  			in = scan.readLine().split(" ");
  			long n = parse(in[0]), m = parse(in[1]);
  			
  			long ans = solve(n, m);
  			out.println(ans);
  		}
  
  		out.flush();
  	}
  
  	private static long solve(long n, long m) 
  	{
  		if(m < n) return 0;
  		
  		long candidate = 0, bit = 1;
  		
  		while(candidate <= m)
  		{
  			candidate |= bit;
  			bit *= 2;
  		}
  		
  		bit /= 2;
  		
  		for( ; bit > 0; bit /= 2)
  		{
  			if((bit & n) > 0) continue; // leave off
  			
  			if((candidate & (~bit)) > m)
  			{
  				candidate ^= bit;
  			}
  		}
  		
  		return n ^ candidate;
  	}
  
  	public static int parse(String num)
  	{
  		return Integer.parseInt(num);
  	}
  }
  