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: import java.util.*;
  import java.io.*;
  
  public class C1557 {
  	static int n, m;
  	static int memo[][];
  
  	public static int solve(int idx, int greater) {
  		if (idx == -1) {
  			return greater == 1 ? 0 : 1 << 30;
  		}
  		if (memo[idx][greater] != -1) {
  			return memo[idx][greater];
  		}
  		int nBit = (n & (1 << idx)) == 0 ? 0 : 1;
  		int mBit = (m & (1 << idx)) == 0 ? 0 : 1;
  		if (greater == 1) {
  			return memo[idx][greater] = solve(idx - 1, 1);
  		}
  		if (mBit == 0) {
  			return memo[idx][greater] = Math.min(solve(idx - 1, 0) + ((nBit ^ 0) << idx),
  					solve(idx - 1, 1) + ((nBit ^ 1) << idx));
  		} else {
  			return memo[idx][greater] = solve(idx - 1, 0) + ((nBit ^ 1) << idx);
  		}
  	}
  
  	public static void main(String[] args) throws IOException {
  		Scanner sc = new Scanner(System.in);
  		PrintWriter pw = new PrintWriter(System.out);
  		int t = sc.nextInt();
  		while (t-- > 0) {
  			n = sc.nextInt();
  			m = sc.nextInt();
  			memo = new int[30][2];
  			for (int[] x : memo)
  				Arrays.fill(x, -1);
  			pw.println(solve(29, 0));
  		}
  		pw.close();
  	}
  
  	static class Scanner {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public Scanner(InputStream s) {
  			br = new BufferedReader(new InputStreamReader(s));
  		}
  
  		public Scanner(FileReader f) {
  			br = new BufferedReader(f);
  		}
  
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  
  		public long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  
  		public double nextDouble() throws IOException {
  			return Double.parseDouble(next());
  		}
  
  		public int[] nextIntArr(int n) throws IOException {
  			int[] arr = new int[n];
  			for (int i = 0; i < n; i++) {
  				arr[i] = Integer.parseInt(next());
  			}
  			return arr;
  		}
  
  	}
  
  }
  