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 Main {
  	public static void main(String[] args) {
  		InputStream inputStream = System.in;
  		OutputStream outputStream = System.out;
  		InputReader in = new InputReader(inputStream);
  		PrintWriter out = new PrintWriter(outputStream);
  		Task solver = new Task();
  		int tcCnt = in.nextInt();
  		for (int tc = 1; tc <= tcCnt; tc++)
  			solver.solve(tc, in, out);
  		out.close();
  	}
  
  	static class Task {
  		public void solve(int testNumber, InputReader in, PrintWriter out) {
  			int n = in.nextInt();
  			int m = in.nextInt();
  			int p = m + 1;
  			int ans = 0;
  			for (int i = 29; i >= 0; i--) {
  				int ni = bit(n, i);
  				int pi = bit(p, i);
  				if (ni == 0 && pi == 1) {
  					ans += 1 << i;
  				} else if(ni == 1 && pi == 0) {
  					break;
  				}
  			}
  			out.println(ans);
  		}
  		int bit(int n, int i) {
  			return (n>>i) & 1;
  		}
  
  	}
  
  	static class InputReader {
  		public BufferedReader reader;
  		public StringTokenizer tokenizer;
  
  		public InputReader(InputStream stream) {
  			reader = new BufferedReader(new InputStreamReader(stream), 32768);
  			tokenizer = null;
  		}
  
  		public String nextLine() {
  			try {
  				return reader.readLine();
  			} catch (IOException e) {
  				throw new RuntimeException(e);
  			}
  		}
  
  		public String nextToken() {
  			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
  				tokenizer = new StringTokenizer(nextLine());
  			}
  			return tokenizer.nextToken();
  		}
  
  		public int nextInt() {
  			return Integer.parseInt(nextToken());
  		}
  
  		public double nextDouble() {
  			return Double.parseDouble(nextToken());
  		}
  
  		public long nextLong() {
  			return Long.parseLong(nextToken());
  		}
  
  		public int[] nextIntArray(int size) {
  			int[] arr = new int[size];
  			for (int i = 0; i < size; i++) {
  				arr[i] = nextInt();
  			}
  			return arr;
  		}
  
  	}
  }