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.util.PriorityQueue;

public class Main{
	static int[] convertIntToBits(int x){
		int [] bitRep = new int[32];
		int i=0;
		while(x!=0){
			bitRep[i] = x&1;
			x = x>>1;
			i++;
		}
		return bitRep;
	}
	static int convertBitToInt(int [] x){
		int ret= 0;
		for(int i=31;i>=0;i--){
			ret = ret<<1;
			ret |= x[i];
		}
		return ret;
	}
	public static void main (String args[]) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int numTc = Integer.parseInt(br.readLine());
		for(int tc = 0; tc <numTc; tc++) {
			String input1 []= br.readLine().split(" ");
			
			int n = Integer.parseInt(input1[0]);

			int m = Integer.parseInt(input1[1]);

			if(m<n)
				System.out.println(0);
			else{
				int p = m+1;
				int bitN []= convertIntToBits(n);
				int bitP [] = convertIntToBits(p);
				int ans[] = new int [32];
				for(int i=31;i>=0;i--){
					if(bitN[i]==0 && bitP[i]==1 )
						ans[i] = 1;
					else if(bitN[i]==1 && bitP[i]==0 ){
						ans[i] = 0;
						break;
					}
					else{
						ans[i] = 0;
					}
				

				}
				System.out.println(convertBitToInt(ans));

			}
			
		}
		
	}
}