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: // 05-Sep-2021
  import java.util.*;
  import java.io.*;
  
  public class B {
  	static class FastReader {
  
  		BufferedReader br;
  		StringTokenizer st;
  
  		private 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());
  		}
  
  		int[] nextIntArray(int n) {
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextInt();
  			return a;
  		}
  
  		int[] nextIntArrayOne(int n) {
  			int[] a = new int[n + 1];
  			for (int i = 1; i < n + 1; i++)
  				a[i] = nextInt();
  			return a;
  		}
  
  		long[] nextLongArray(int n) {
  			long[] a = new long[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextLong();
  			return a;
  		}
  
  		long[] nextLongArrayOne(int n) {
  			long[] a = new long[n + 1];
  			for (int i = 1; i < n + 1; i++)
  				a[i] = nextLong();
  			return a;
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		String nextLine() {
  			String str = "";
  			try {
  				str = br.readLine();
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  	}
  
  	public static void main(String[] args) {
  		FastReader s = new FastReader();
  		StringBuilder str = new StringBuilder();
  
  		int t = s.nextInt();
  		int xor[] = new int[(int)(3e5 + 1000)];
  		
  		for(int i = 1; i < xor.length; i++) {
  			xor[i] = i ^ xor[i - 1];
  		}
  		
  		while (t-- > 0) {
  			int a = s.nextInt(), b = s.nextInt();
  			int ans = a;
  			int x = xor[a - 1];
  			
  			if(x != b) {
  				
  				int mask = b ^ x;
  				if(mask == a) {
  					ans +=2;
  				}else ans++;
  			}
  			
  			str.append(ans+"\n");
  		}
  		System.out.println(str);
  	}
  
  }
  