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.*;
  import java.util.*;
  
  public class B {
  	public static void main(String args[]){
  		FScanner in = new FScanner();
  		PrintWriter out = new PrintWriter(System.out);
  		int t = in.nextInt();
  		while(t-->0) { 
  			int a=in.nextInt();
  			int b=in.nextInt();
  			int rem=(a-1)%4;
  
  			if(rem==0)
  			{
  				if(b==a-1)
  					out.println(a);
  				else 
  				{
  					int and=(a-1)^b;
  					if(and>a||and<a)
  						out.println(a+1);
  					else
  						out.println(a+2);
  				}
  //169234 169235
  			}
  
  			if(rem==1)
  			{
  				if(b==1)
  					out.println(a);
  				else
  				{
  					int and=1^b;
  					if(and>a||and<a)
  						out.println(a+1);
  					else
  						out.println(a+2);
  				}
  			}
  
  			if(rem==2)
  			{
  				if(b==a)
  					out.println(a);
  				else 
  				{
  					int and=a^b;
  					if(and>a||and<a)
  						out.println(a+1);
  					else
  						out.println(a+2);
  				}
  			}
  
  			if(rem==3)
  			{
  				if(b==0)
  					out.println(a);
  				else
  				{
  					int and =0^b;
  					if(and<a||and>a)
  						out.println(a+1);
  					else
  						out.println(a+2);
  				}
  			}
  
  
  
  		}
  		//out.println(1^169235);
  		out.close();
  	}
  
  	static boolean checkprime(int n1)
  	{
  		if(n1%2==0||n1%3==0)
  			return false;
  		else 
  		{
  			for(int i=5;i*i<=n1;i+=6)
  			{
  				if(n1%i==0||n1%(i+2)==0)
  					return false;
  			}
  			return true;
  		}
  	}
  	
  	static class FScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer sb = new StringTokenizer("");
  
  		String next(){
  			while(!sb.hasMoreTokens()){
  				try{
  					sb = new StringTokenizer(br.readLine());
  				} catch(IOException e){ }
  			}
  			return sb.nextToken();
  		}
  		String nextLine(){
  			try{ return br.readLine(); } 
  			catch(IOException e) { } return "";
  		}
  
  		int nextInt(){
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		int[] readArray(int n) {
  			int a[] = new int[n];
  			for(int i=0;i<n;i++) a[i] = nextInt();
  			return a;
  		}
  
  		float nextFloat(){
  		return Float.parseFloat(next());
  		}
  		
  		double nextDouble(){
  		return Double.parseDouble(next());
  		}
  	}
  }
  