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.

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);
		
		*/
		for (int i = 0; ;i++) {
			if ((i ^ n) > m) {
				System.out.println(i);
				break;
			}
		}
		
		
		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;
		}
	}
}
