description: 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.
solution: import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.text.DecimalFormat;
  import java.util.*;
  
  //import sun.java2d.opengl.WGLSurfaceData.WGLVSyncOffScreenSurfaceData;
  
  
   
  public class Codeforces {
  	
  	static int mod=1000000007 ;
  
  	public static void main(String[] args) throws Exception {
  		PrintWriter out=new PrintWriter(System.out);
  		FastScanner fs=new FastScanner();
  		int t=fs.nextInt();
  		while(t-->0) {
  			long x=fs.nextLong(), y=fs.nextLong();
  			long ans=0;
  			if(y<x) {
  				ans=x+y;
  			}
  			else if(x==y) {
  				ans= x;
  			}
  			else {
  				if(y%x==0) {
  					ans=y;
  				}
  				else {
  					long n= (y/x)*x;
  					
  					long mid= n+ (y-n)/2;
  					ans=mid;
  				}
  			}
  //			if(ans%x!=y%ans) System.out.println("false");
  			
  			// n % x == y % n
  			out.println(ans);
  		}
  		out.close();
  		
  	}
  	
  	static long pow(long a,long b) {
  		if(b<0) return 1;
  		long res=1;
  		while(b!=0) {
  			if((b&1)!=0) {
  				res*=a;
  				res%=mod;
  			}
  			a*=a;
  			a%=mod;
  			b=b>>1;
  		}
  		return res;
  	}
  	static long gcd(long  a,long  b) {
  		if(b==0) return a;
  		return gcd(b,a%b);
  	}
  	static long nck(int n,int k) {
  		if(k>n) return 0;
  		long res=1;
  		res*=fact(n);
  		res%=mod;
  		res*=modInv(fact(k));
  		res%=mod;
  		res*=modInv(fact(n-k)); 
  		res%=mod;
  		return res;
  	}
  	static long fact(long n) {
  //		return fact[(int)n];
  		long res=1;
  		for(int i=2;i<=n;i++) {
  			res*=i;
  			res%=mod;
  		}
  		return res;
  	}
  	
  	static long modInv(long n) {
  		return pow(n,mod-2);
  	}
  	
  	static void sort(int[] a) {
  		//suffle
  		int n=a.length;
  		Random r=new Random();
  		for (int i=0; i<a.length; i++) {
  			int oi=r.nextInt(n);
  			int temp=a[i];
  			a[i]=a[oi];
  			a[oi]=temp;
  		}
  		
  		//then sort
  		Arrays.sort(a);
  	}
  	
  	// Use this to input code since it is faster than a Scanner
  	static class FastScanner {
  		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());
  		}
  		long[] lreadArray(int n) {
  			long a[]=new long[n];
  			for(int i=0;i<n;i++) a[i]=nextLong();
  			return a;
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  	}
   
  }