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 Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		Task solver = new Task();
		int tcCnt = in.nextInt();
		for (int tc = 1; tc <= tcCnt; tc++)
			solver.solve(tc, in, out);
		out.close();
	}

	static class Task {
		public void solve(int testNumber, InputReader in, PrintWriter out) {
			int n = in.nextInt();
			int m = in.nextInt();
			int p = m + 1;
			int ans = 0;
			for (int i = 29; i >= 0; i--) {
				int ni = bit(n, i);
				int pi = bit(p, i);
				if (ni == 0 && pi == 1) {
					ans += 1 << i;
				} else if(ni == 1 && pi == 0) {
					break;
				}
			}
			out.println(ans);
		}
		int bit(int n, int i) {
			return (n>>i) & 1;
		}

	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}

		public String nextToken() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				tokenizer = new StringTokenizer(nextLine());
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(nextToken());
		}

		public double nextDouble() {
			return Double.parseDouble(nextToken());
		}

		public long nextLong() {
			return Long.parseLong(nextToken());
		}

		public int[] nextIntArray(int size) {
			int[] arr = new int[size];
			for (int i = 0; i < size; i++) {
				arr[i] = nextInt();
			}
			return arr;
		}

	}
}