YouKn0wWho has two even integers x and y. Help him to find an integer n such that 1 ≤ n ≤ 2 ⋅ 10^{18} and n mod x = y mod n. Here, a mod b denotes the remainder of a after division by b. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.

The first and only line of each test case contains two integers x and y (2 ≤ x, y ≤ 10^9, both are even).

Output

For each test case, print a single integer n (1 ≤ n ≤ 2 ⋅ 10^{18}) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints.

Example

Input


4
4 8
4 2
420 420
69420 42068


Output


4
10
420
9969128

Note

In the first test case, 4 mod 4 = 8 mod 4 = 0.

In the second test case, 10 mod 4 = 2 mod 10 = 2.

In the third test case, 420 mod 420 = 420 mod 420 = 0.
1
1000000000 1000000000

1000000000 

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.out;
import java.util.*;
import java.io.File;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.math.BigInteger;
public class Main {
 
	
	/* 10^(7) = 1s.
	 * ceilVal = (a+b-1) / b */
	
	static final int mod = 1000000007;
	static final long temp = 998244353;
	static final long MOD = 1000000007;
	static final long M = (long)1e9+7;
 
	static class Pair implements Comparable<Pair> 
	{
		int first, second;
		
		public Pair(int first, int second) 
		{
			this.first = first;
			this.second = second;
		}
		
		public int compareTo(Pair ob) 
		{
			return (int)(first - ob.first);
		}
	}
	
	
	static class Tuple implements Comparable<Tuple>
	{
		int first, second,third;
		public Tuple(int first, int second, int third)
		{
			this.first = first;
			this.second = second;
			this.third = third;
		}
		public int compareTo(Tuple o)
		{
			return (int)(o.third - this.third);
		}
	}
	
	public static class DSU
	{
		int[] parent;
		int[] rank; //Size of the trees is used as the rank
		public DSU(int n)
		{
			parent = new int[n];
			rank = new int[n];
			Arrays.fill(parent, -1);
			Arrays.fill(rank, 1);
		}
		
		public int find(int i) //finding through path compression
		{
			return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
		}
		
		public boolean union(int a, int b) //Union Find by Rank 
		{
			a = find(a);
			b = find(b);
			
			if(a == b) return false; //if they are already connected we exit by returning false.
			
			// if a's parent is less than b's parent
			if(rank[a] < rank[b])
			{
				//then move a under b
				parent[a] = b;
			}
			//else if rank of j's parent is less than i's parent
			else if(rank[a] > rank[b])
			{
				//then move b under a
				parent[b] = a;
			}
			//if both have the same rank.
			else
			{
				//move a under b (it doesnt matter if its the other way around.
				parent[b] = a;
				rank[a] = 1 + rank[a];
			}
			return true;
		}	
	}

	static class Reader 
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		int[] readArray(int n) throws IOException {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		long[] longReadArray(int n) throws IOException {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	
	public static int gcd(int a, int b)
	{
		if(b == 0)
		 return a;
		
		else
		return gcd(b,a%b);
	} 
 
		public static long lcm(long a, long b)
	    {
	        return (a / LongGCD(a, b)) * b;
	    }
	
	public static long LongGCD(long a, long b)
	{
		if(b == 0)
			 return a;
			
			else
			return LongGCD(b,a%b);
	}
	
		public static long LongLCM(long a, long b)
	    {
	        return (a / LongGCD(a, b)) * b;
	    }
	
		//Count the number of coprime's upto N
		public static long phi(long n)  //euler totient/phi function
	    { 
	       long ans = n;
//	       for(long i = 2;i*i<=n;i++)
//	       {
//	    	   if(n%i == 0)
//	    	   {
//	    		   while(n%i == 0) n/=i;
//	    		   ans -= (ans/i);
//	    	   }
//	       }
//	       
//	       if(n > 1)
//	       {
//	    	   ans -= (ans/n);
//	       }
	       
	       for(long i = 2;i<=n;i++)
	       {
	    	   if(isPrime(i))
	    	   {
	    		   ans -= (ans/i);
	    	   }
	       }
	       
	       return ans;
	       
	    }	
	
	public static long fastPow(long x, long n) 
	{
		if(n == 0)
			return 1;
		else if(n%2 == 0)
			return fastPow(x*x,n/2);
		else
			return x*fastPow(x*x,(n-1)/2);
	}
	
	   public static long modPow(long x, long y, long p)
	    {
	        long res = 1;
	        x = x % p;
	 
	        while (y > 0) {
	            if (y % 2 == 1)
	                res = (res * x) % p;
	 
	            y = y >> 1; 
	            x = (x * x) % p;
	        }
	        return res;
	    }
	   
	   static long modInverse(long n, long p)
	    {
	        return modPow(n, p - 2, p);
	    }
	 
	    // Returns nCr % p using Fermat's little theorem.
	    
	    public static long nCrModP(long n, long r,long p)
	    {
	    	if (n<r) 
	              return 0;
	          
	        if (r == 0)
	            return 1;
	 
	        long[] fac = new long[(int)(n) + 1];
	        fac[0] = 1;
	 
	        for (int i = 1; i <= n; i++)
	            fac[i] = fac[i - 1] * i % p;
	 
	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
	                % p * modInverse(fac[(int)(n - r)], p)
	                % p)
	            % p;
	    }
	    
	    public static long fact(long n) 
	    {
	    	  long[] fac = new long[(int)(n) + 1];
		        fac[0] = 1;
		 
		        for (long i = 1; i <= n; i++)
		            fac[(int)(i)] = fac[(int)(i - 1)] * i;
		        
		        return fac[(int)(n)];
		}
	    
	    public static long nCr(long n, long k) 
	    {
			long ans = 1;
			for(long i = 0;i<k;i++)
			{
				ans *= (n-i);
				ans /= (i+1);
			}
			return ans;
		}
 
		//Modular Operations for Addition and Multiplication. 
		   public static long perfomMod(long x)
		   {
		        return ((x%M + M)%M);
		    }
		   public static long addMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)+perfomMod(b));
		    }
		  
		   public static long subMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)-perfomMod(b));
		   }
		   
		   public static long mulMod(long  a, long b)
		   {
		        return perfomMod(perfomMod(a)*perfomMod(b));
		   }
		   
		   public static boolean isPrime(long n) 
			{
				if(n == 1)
				{
					return false;
				}
				
				//check only for sqrt of the number as the divisors 
				//keep repeating so only half of them are required. So,sqrt.
				for(int i = 2;i*i<=n;i++)
				{
					if(n%i == 0)
					{
						return false;
					}
				}
				return true;
			}
			
			public static List<Long> SieveList(int n)
			{
				boolean prime[] = new boolean[(int)(n+1)];
				Arrays.fill(prime, true);
				
				List<Long> l = new ArrayList<>();
				for (long p = 2; p*p<=n; p++) 
				{ 
					if (prime[(int)(p)] == true) 
					{ 	
					    for(long i = p*p; i<=n; i += p) 
						{
						    prime[(int)(i)] = false; 
						}	
					} 
				} 
		 
				for (long p = 2; p<=n; p++) 
				{
				    if (prime[(int)(p)] == true)
				    {
				       l.add(p); 
				    }
				}
				
				return l;
			}

			public static long log2(long n)
		    {
				 long ans = (long)(log(n)/log(2));
				 return ans;
			}
			   
			public static boolean isPow2(long n)
			{
				 return (n  != 0 && ((n & (n-1))) == 0);
			}	
		   
		   public static boolean isSq(int x)  
		   {
			    long s = (long)Math.round(Math.sqrt(x));
			    return s*s==x;
		   }
		
			   /*
			    * 
			    * >=         <=
			   0 1 2 3 4 5 6 7  
			   5 5 5 6 6 6 7 7
			   lower_bound for 6 at index 3 (>=)
			   upper_bound for 6 at index 6(To get six reduce by one) (<=)
			   */

		   public static int LowerBound(long a[], long x) 
		   {
			      int l=-1,r=a.length;
				  while(l+1<r) 
				  {
				    int m=(l+r)>>>1;
				    if(a[m]>=x) r=m;
				    else l=m;
				  }
				  return r;
			}
		   
			public static int UpperBound(long a[], long x) 
			{
				int l=-1, r=a.length;
			    while(l+1<r) 
			    {
			       int m=(l+r)>>>1;
			       if(a[m]<=x) l=m;
			       else r=m;
			    }
			    return l+1;
			}
			
			public static void Sort(long[] a) 
			{
				List<Long> l = new ArrayList<>();
				for (long i : a) l.add(i);
				Collections.sort(l);
				Collections.reverse(l);  //Use to Sort decreasingly 
				for (int i=0; i<a.length; i++) a[i]=l.get(i);
			}
			
			public static void ssort(char[] a)
			{
				List<Character> l = new ArrayList<>();
				for (char i : a) l.add(i);
				Collections.sort(l);
				for (int i=0; i<a.length; i++) a[i]=l.get(i);
			}
		
			
	public static void main(String[] args) throws Exception 
	{
	
		Reader sc = new Reader();
		PrintWriter fout = new PrintWriter(System.out);

		int t = sc.nextInt();
		while(t-- > 0)
		{
			long x = sc.nextLong(), y = sc.nextLong();
			
			if(x == y)
			{
				fout.println(x);
			}
			else if(x > y)
			{
				fout.println((x + y));
			}
			else 
			{
				if(y%x == 0)
				{
					fout.println(x);
				}
				else
				{
					long rem=  y%x;
					fout.println((y - rem/2));
				}
			}
		}
		
		fout.close();
	}
}