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.Scanner;

public class CF735C {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s= new Scanner(System.in);
		int t=s.nextInt();
		while(t-->0) {
			int n=s.nextInt();
			int m=s.nextInt();
			solve(n,m);
		}

	}
	static void solve(int n,int m) {
//		System.out.println(n+" "+m);
		int bit=32;
		int ans=0;
		//n^k>=(m+1)
		for(int i = bit-1;i>=0;i--) {
			//n=1 and m+1 =1
			int a=n&(1<<i);
			int b=(1<<i)&(m+1);
//			System.out.println("i:"+i+" a:"+a+" b:"+b);
			if(a>=1 && b>=1){
				
			}
			//n=0 and m+1 =1
			else if(a==0 && b>=1){
//				System.out.println("here");
				ans+=(1<<i);
			}
			//n=1 and m+1 =0
			else if(a>=1 && b==0){
				break;
			}
			//n=0 and m+1 =0
			else{
				
			}
		}
		System.out.println(ans);
	}

}
