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.


import java.util.*;
import java.io.*;
import java.math.BigInteger;

public class P04 {
	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 void swap(int[] arr, int i, int j)
	{
	    int temp = arr[i];
	    arr[i] = arr[j];
	    arr[j] = temp;
	}

	static int partition(int[] arr, int low, int high)
	{	   
	    long pivot = arr[high];   
	    int i = (low - 1);
	    for(int j = low; j <= high - 1; j++)
	    {    
	        if (arr[j] < pivot)
	        {          
	            i++;
	            swap(arr, i, j);
	        }
	    }
	    swap(arr, i + 1, high);
	    return (i + 1);
	}
	static void quickSort(int[] arr, int low, int high)
	{
	    if (low < high)
	    {  
	        int pi = partition(arr, low, high);
	        quickSort(arr, low, pi - 1);
	        quickSort(arr, pi + 1, high);
	    }
	}
	public static long binarySearch(long arr[], long key){  
		  int  first =0,last =arr.length-1;
		   int mid = (first + last)/2;  
		   while( first <= last ){  
		      if ( arr[mid] < key ){  
		        first = mid + 1;     
		      }else if ( arr[mid] == key ){  
		    	 return mid;  
		      }else{  
		         last = mid - 1;  
		      }  
		      mid = (first + last)/2;  
		   }  
		   if ( first > last ){ 
			   return -1;
		       
		   } 
		   return -1;
		 }  
	static long fp(long a , long b , int n) {
		long res =1;
		while(b>0) {
			if((b&1)!=0) {             //when b is odd 
				res=(res*a%n)%n;
			}
			a=(a%n*a%n)%n;
			b=b>>1;
		}
		return res;
	}
	static long gcd (long a ,long b) {
		if(b==0) {
			return a;
		}
		return gcd(b,a%b);
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner (System.in);
		FastReader fs = new FastReader();
		int sr []= new int [300001];
		sr[1]=0;
		for(int i =2;i<300001;i++) {
			sr[i]=sr[i-1]^i-1;
		}
		int t = fs.nextInt();
		while(t-->0) {
			int a = fs.nextInt();
			int b = fs.nextInt();
			if(sr[a]==b)System.out.println(a);
			else if(b%2==0||sr[a]==0) {
				if(a==b)System.out.println(a+2);
				else System.out.println(a+1);
			}
			else System.out.println(a+2);
		}
		
    }
	

	private static String reduce(String s) {
		String ans ="";
		ans+=s.charAt(0);
		for(int i =1;i<s.length();i++) {
			if(s.charAt(i)!=s.charAt(i-1)) {
				ans+=s.charAt(i);
			}
		}
	//	if(s.charAt(s.length()-1)!=s.charAt(s.length()-2))ans+=s.charAt(s.length()-1);
		return ans;
	}

	private static int solve(int[] a,int b[], int n) {
		int flag =0;
		for(int j =0;j<n;j++) {
			if(a[j]!=b[j]) {
				flag=1;
			}
		}
		if(flag==0)return 0;
		int i=1;
		while(true) {
			if(i%2==0) {
				for(int j=1;j<n-1;j+=2) {
					if(a[j]>a[j+1]) {
						int temp =a[j];
						a[j]=a[j+1];
						a[j+1]=temp;
					}
				}
			}
			else {
				for(int j =0;j<n-2;j+=2) {
					if(a[j]>a[j+1]) {
						int temp =a[j];
						a[j]=a[j+1];
						a[j+1]=temp;
					}
				}
			}
			flag =0;
			for(int j =0;j<n;j++) {
				if(a[j]!=b[j])flag=1;
			}
			if(flag==0)return i;
			i++;
		}
       
		
		
	}

	private static long sumofn(long n) {
		int ans = 0;
		while(n>9) {
			ans+=n%10;
			n/=10;
		}
		ans+=n;
		return ans;
	}
	private static boolean palin(String s) {
		int i =0,j=s.length()-1;
		while(i<j) {
			if(s.charAt(i)!=s.charAt(j))return false;
			else {
				i++;
				j--;
			}
		}
		return true;
	}
}
			


