description: Alice gave Bob two integers a and b (a > 0 and b ≥ 0). Being a curious boy, Bob wrote down an array of non-negative integers with \operatorname{MEX} value of all elements equal to a and \operatorname{XOR} value of all elements equal to b.
  
  What is the shortest possible length of the array Bob wrote?
  
  Recall that the \operatorname{MEX} (Minimum EXcluded) of an array is the minimum non-negative integer that does not belong to the array and the \operatorname{XOR} of an array is the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all the elements of the array.
  
  Input
  
  The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 5 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
  
  The only line of each test case contains two integers a and b (1 ≤ a ≤ 3 ⋅ 10^5; 0 ≤ b ≤ 3 ⋅ 10^5) — the \operatorname{MEX} and \operatorname{XOR} of the array, respectively.
  
  Output
  
  For each test case, output one (positive) integer — the length of the shortest array with \operatorname{MEX} a and \operatorname{XOR} b. We can show that such an array always exists.
  
  Example
  
  Input
  
  
  5
  1 1
  2 1
  2 0
  1 10000
  2 10000
  
  
  Output
  
  
  3
  2
  3
  2
  3
  
  Note
  
  In the first test case, one of the shortest arrays with \operatorname{MEX} 1 and \operatorname{XOR} 1 is [0, 2020, 2021].
  
  In the second test case, one of the shortest arrays with \operatorname{MEX} 2 and \operatorname{XOR} 1 is [0, 1].
  
  It can be shown that these arrays are the shortest arrays possible.
solution: /*
                         _oo0oo_
                        o8888888o
                        88" . "88
                        (| -_- |)
                        0\  =  /0
                      ___/`---'\___
                    .' \\|     |// '.
                   / \\|||  :  |||// \
                  / _||||| -:- |||||- \
                 |   | \\\  -  /// |   |
                 | \_|  ''\---/''  |_/ |
                 \  .-\__  '-'  ___/-. /
               ___'. .'  /--.--\  `. .'___
            ."" '<  `.___\_<|>_/___.' >' "".
           | | :  `- \`.;`\ _ /`;.`/ - ` : | |
           \  \ `_.   \_ __\ /__ _/   .-` /  /
       =====`-.____`.___ \_____/___.-`___.-'=====
                         `=---='
   
  */
  import java.util.function.Consumer;
  import java.util.*;
  import java.math.*;
  import java.io.*;
  import java.lang.Math.*;
  
  public class KickStart2020 {
  	static class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		String next() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		float nextFloat() {
  			return Float.parseFloat(next());
  		}
  
  		String nextLine() {
  			String str = "";
  			try {
  				str = br.readLine();
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  	}
  
  
  	static int computeXOR(int n)
      {
          // If n is a multiple of 4
          if (n % 4 == 0)
              return n;
        
          // If n%4 gives remainder 1
          if (n % 4 == 1)
              return 1;
        
          // If n%4 gives remainder 2
          if (n % 4 == 2)
              return n + 1;
        
          // If n%4 gives remainder 3
          return 0;
      }
  	public static void main(String[] args) throws Exception {
  		FastReader sc = new FastReader();
  		PrintWriter out = new PrintWriter(System.out);
  		int t = sc.nextInt();
  		outerloop:
  		while(t-- > 0) {
              int n = sc.nextInt();
              int x = sc.nextInt();
              int sum = n;
              int xor = computeXOR(n - 1);
              if((xor ^ x) == n) sum += 2;
              else if(xor != x) sum++;
              out.println(sum);
  		}
  		out.close();
  	}
  }