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 GFG {
    
    public static class Pair{
        long x;
        int y;
        Pair(long x,int y){
            this.x = x;
            this.y = y;
        }
    }
    
	public static void main (String[] args) {
	    Scanner sc = new Scanner(System.in);
	    int t = sc.nextInt();
	     while(t-- > 0){
	         long n = sc.nextLong();
	         long m = sc.nextLong();
	         long prev = 0,ans = Long.MAX_VALUE;
	         for(long i=32;i>=0;i--)
	         {
	             long bit1 = (1L<<i)&n;
	             long bit2 = (1L<<i)&m;
	             if(bit2>0){
	                 if(bit1==0)
	                 prev |= (1L<<i);
	             }
	             else{
	                 if(bit1==0)
	                 ans = Math.min(ans,prev|(1L<<i));
	                 else{
	                     ans = Math.min(ans,prev);
	                     prev|=bit1;
	                 }
	             }
	         }
	         System.out.println(ans);
	     }    
	}
}
