1554_C. Mikasa_122
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.*;
import java.io.*;

public class c {

	public static void main(String[] args) throws Exception {
	
		BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer tok = new StringTokenizer(stdin.readLine());
		int nC = Integer.parseInt(tok.nextToken());
		
		StringBuffer sb = new StringBuffer();
		for (int loop=0; loop<nC; loop++) {
			tok = new StringTokenizer(stdin.readLine());
			int n = Integer.parseInt(tok.nextToken());
			int m = Integer.parseInt(tok.nextToken());
			sb.append(solve(n,m)+"\n");
			//System.out.println(solve(n,m)+" "+altSolve(n,m));
		}
		System.out.print(sb);
	}
	
	public static int altSolve(int n, int m) {
		
		int next = numBits(Math.max(n,m))+1;
		boolean[] used = new boolean[1<<next];
		
		for (int i=0; i<=m; i++)
			used[n^i] = true;
		for (int i=0; i<used.length; i++)
			if (!used[i])
				return i;
		return -1;
		
	}
	
	public static int solve(int n, int m) {
		
		if (m < n) return 0;
		
		// all ones case.
		if (perfect2(m)) return m+1;
		
		// cases same bits
		// m has more bits than n.
		int mBits = numBits(m);
		int nBits = numBits(n);
		
		if (mBits > nBits) {
			int restM = (m >> nBits) << nBits;
			int add = solve(n, m & ((1<<nBits)-1));
			return restM+add;
		}
		
		//System.out.println(Integer.toBinaryString(n));
		//System.out.println(Integer.toBinaryString(m));

		int lastZeroBit = 1;
		while ((m & lastZeroBit) != 0) lastZeroBit <<= 1;
		
		int start = Integer.highestOneBit(n);
		boolean hasOneZero = false;
		int res = 0;
		
		for (int i=start; i>=1; i>>=1) {
			int mChoice = m & i;
			int nChoice = n & i;
			
			if (nChoice == 0 && mChoice > 0)
				res += i;
			else if (nChoice > 0 && mChoice > 0 && i < lastZeroBit)
				res += i;
			else if (nChoice > 0 && mChoice == 0) {
				hasOneZero = true;
				break;
			}
		}
		

		
		if (hasOneZero) return res;
		else {
			
			return res + 1;
		}
	}
	
	public static boolean perfect2(int m) {
		return Integer.bitCount(m+1) == 1;
	}
	
	public static int numBits(int n) {
		int res = 0;
		while (n > 0) {
			res++;
			n /= 2;
		}
		return res;
	}
}