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.*;

public class Practice {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-->0) {
            String a = (Integer.toBinaryString(sc.nextInt()));
            String b = (Integer.toBinaryString(sc.nextInt() + 1));
            int[] val1 = new int[32];
            int[] val2 = new int[32];
            int index = 31;
            for (int i = a.length() - 1; i >= 0; i--) {
                val1[index] = a.charAt(i) - '0';
                index--;
            }
            index = 31;
            for (int i = b.length() - 1; i >= 0; i--) {
                val2[index] = b.charAt(i) - '0';
                index--;
            }
            StringBuilder ans = new StringBuilder();
            for (int i = 0; i < 32; i++) {
                if (val1[i] == val2[i]) ans.append(0);
                else if (val1[i] == 1 && val2[i] == 0) {
                    while (i < 32) {
                        ans.append(0);
                        i++;
                    }
                } else {
                    ans.append(1);
                }
            }
            System.out.println(Integer.parseInt(ans.toString(), 2));
        }
    }
}

