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.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);
	}
}
