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: import java.io.BufferedReader;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  
  public class taskb {
  	public static void main(String[] args) {
  		FastScanner in = new FastScanner();
  		PrintWriter out = new PrintWriter(System.out);
  		int xor_prefix[] = new int[300100];
  		for (int i = 1; i <= (int) 3e5; i++) {
  			xor_prefix[i] = xor_prefix[i-1] ^ i;
  		}
  		int tt = in.nextInt();
  		for (int tc = 0; tc < tt; tc++) {
  			int mex = in.nextInt(), xor = in.nextInt();
  			int cur_xor = xor_prefix[mex-1];
  			if (cur_xor == xor) {
  				out.println(mex);
  			} else {
  				int need = xor ^ cur_xor;
  				if (need == mex) {
  					out.println(mex + 2);
  				} else {
  					out.println(mex + 1);
  				}
  			}
  		}
  		out.flush();
  	}
  
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer tok = new StringTokenizer("");
  		String next() {
  			while(!tok.hasMoreTokens()) {
  				try {
  					tok = new StringTokenizer(br.readLine());
  				} catch (Exception e) {
  					e.printStackTrace();
  				}
  			}
  			return tok.nextToken();
  		}
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		float nextFloat() {
  			return Float.parseFloat(next());
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  }
  