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++;
    					r|=n;
    					if((r^n)>r)
    					{
    						System.out.println(r);
    					}
    					else
    					{
    						System.out.println(r^n);
    					}
    				}
    				else
    				{
    					System.out.println(r^n);
    				}
    			}
    		}
    		sc.close();
    	}
    }