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.*;
public class Solution
{
	public static void main(String args[])
	{
		Scanner sc=new Scanner(System.in); 
		int test= sc.nextInt();
		for(int xix=1;xix<=test;xix++)
		{
			int n=sc.nextInt(); int m=sc.nextInt();
			if(n>m)
			{
				System.out.println(0);
			}
			else
			{
				String q=Integer.toBinaryString(((n^m)&n));
				String s=Integer.toBinaryString(m);
				String ans=s.substring(0,s.length()-q.length())+q;
				if(q.equals("0"))
				{
					ans=s;
				}
				int r=Integer.parseInt(ans,2);
				r|=n;
				if(r==m)
				{
					r++;
					if((r^n)>r)
					{
						System.out.println(r);
					}
					else
					{
						System.out.println(r^n);
					}
				}
				else
				{
					System.out.println(r^n);
				}
			}
		}
		sc.close();
	}
}