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 CF735C {
  	public static void solve(IO io) {
  
  		long n = io.getLong();
  		long m = io.getLong();
  	/*	int m = (int) (Math.random()*1000000000);
  		int n = (int) (Math.random()*m);
  		
  		*/
  
  		
  		
  		if (n > m) {
  			System.out.println(0);
  			return;
  		}
  		
  		if (n == 0) {
  			System.out.println(m + 1);
  			return;
  		}
  		
  		String ns = Long.toBinaryString(n);
  		String ms = Long.toBinaryString(m + 1);
  
  		for (int i = ns.length(); i < ms.length(); i++) {
  			ns = "0" + ns;
  		}
  		
  		String ans = (ns.charAt(0) == '1')? "0":"1";
  		boolean lol = false;
  		
  		for (int i = 1; i < ns.length(); i++) {
  			if (ns.charAt(i) == ms.charAt(i) || lol) {
  				ans += "0";
  			} else if (ns.charAt(i) == '0') {
  				ans += "1";
  			} else {
  				ans += "0";
  				lol = true;
  			}
  		}
  		
  		System.out.println(Long.parseLong(ans,2));
  	//	System.out.println(ns);
  	//	System.out.println(ms);
  	//	System.out.println(ans);
  	//	System.out.println();
  
  	}
  
  	public static void main(String[] args) {
  		IO io = new IO();
  		int t = io.getInt();
  		for (int testIndex = 0; testIndex < t; testIndex++) {
  			solve(io);
  		}
  		io.close();
  	}
  
  	static class IO extends PrintWriter {
  
  		public IO() {
  			super(new BufferedOutputStream(System.out));
  			r = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		public IO(InputStream i) {
  			super(new BufferedOutputStream(System.out));
  			r = new BufferedReader(new InputStreamReader(i));
  		}
  
  		public IO(InputStream i, OutputStream o) {
  			super(new BufferedOutputStream(o));
  			r = new BufferedReader(new InputStreamReader(i));
  		}
  
  		public boolean hasMoreTokens() {
  			return peekToken() != null;
  		}
  
  		public int getInt() {
  			return Integer.parseInt(nextToken());
  		}
  
  		public double getDouble() {
  			return Double.parseDouble(nextToken());
  		}
  
  		public long getLong() {
  			return Long.parseLong(nextToken());
  		}
  
  		public String getWord() {
  			return nextToken();
  		}
  
  		private BufferedReader r;
  		private String line;
  		private StringTokenizer st;
  		private String token;
  
  		private String peekToken() {
  			if (token == null)
  				try {
  					while (st == null || !st.hasMoreTokens()) {
  						line = r.readLine();
  						if (line == null)
  							return null;
  						st = new StringTokenizer(line);
  					}
  					token = st.nextToken();
  				} catch (IOException e) {
  				}
  			return token;
  		}
  
  		private String nextToken() {
  			String ans = peekToken();
  			token = null;
  			return ans;
  		}
  	}
  }
  