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.
// coached by kaiboy and rainboy
import java.io.*;
import java.util.*;

public class CF1554C extends PrintWriter {
	CF1554C() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1554C o = new CF1554C(); o.main(); o.flush();
	}

	static final int K = 31;
	void decode(int[] aa, int a) {
		for (int h = 0; h < K; h++)
			aa[h] = a >> h & 1;
	}
	int encode(int[] aa) {
		int a = 0;
		for (int h = 0; h < K; h++)
			a |= aa[h] << h;
		return a;
	}
	void main() {
		int[] aa = new int[K];
		int[] bb = new int[K];
		int t = sc.nextInt();
		while (t-- > 0) {
			int a = sc.nextInt();
			int b = sc.nextInt() + 1;
			if (b <= a) {
				println(0);
				continue;
			}
			decode(aa, a);
			decode(bb, b);
			int h = K - 1;
			while (h >= 0 && bb[h] >= aa[h]) {
				bb[h] ^= aa[h];
				h--;
			}
			while (h >= 0)
				bb[h--] = 0;
			println(encode(bb));
		}
	}
}
