description: 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.
solution: // 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));
  		}
  	}
  }
  