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.Scanner;

public class Main {

    public static void main(String[] agrs) {
        int bit = 31;
        Scanner sc = new Scanner(System.in);
        int testCase = sc.nextInt();
        for(int i=0; i<testCase; i++) {
           int n = sc.nextInt();
           int m = sc.nextInt();
            int ans = 0;
           for(int j=bit-1; j>=0; j--) {
               //n->1 & m+1->1
               if((1<<j & n) != 0 &&  ((1<<j)&(m+1)) != 0) {
               }
               //n->0 & m+1->1
               else if(((1<<j)&n) == 0 && ((1<<j)&(m+1)) > 0) {
                   ans =ans | (1<<j);
               }
               //n->1 & m+1->0
               else if (((1<<j) & n) > 0 && ((1<<j)&(m+1)) == 0) {
                   break;
               }
               //n->0 & m+1->0
               else {

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