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.util.*;
import java.io.*;

import static java.lang.Math.min;
import static java.lang.Math.max;
import static java.lang.Math.abs;
import static java.lang.Math.sqrt;

public class Solution {

	static Scanner scn = new Scanner(System.in);
	static StringBuilder sb = new StringBuilder();
	static PrintWriter out = new PrintWriter(System.out);

	// https://randommashupgenerator.netlify.app/
	// https://a2oj.com/Ladder6.html
	// https://codeforces.com/problemset?order=BY_RATING_ASC&tags=binary+search%2C1400-1400
	// https://codeforces.com/problemset?order=BY_SOLVED_DESC&tags=1500-1500
	// https://www.webtoons.com/en/challenge/cp-series/binary-search/viewer?title_no=365151&episode_no=5
	// https://github.com/ZRonchy/Algorithms

	public static void main(String[] HastaLaVistaLa) {
		int t = scn.nextInt();
		for(int tests = 0; tests < t; tests++) solve1();
		out.println(sb);
		out.close();
	}

	public static void solve1() {
		long n = scn.nextLong(), m = scn.nextLong();
		long min = 0, max = n ^ m + 1;
		while(min < max) {
			long mid = min + (max - min) / 2;
			long xor = n ^ mid;
			if(xor <= m) min = mid + 1;
			else max = mid;
		}
		sb.append(min);
		sb.append("\n");
	}
}