Alice and Borys are playing tennis.

A tennis match consists of games. In each game, one of the players is serving and the other one is receiving.

Players serve in turns: after a game where Alice is serving follows a game where Borys is serving, and vice versa.

Each game ends with a victory of one of the players. If a game is won by the serving player, it's said that this player holds serve. If a game is won by the receiving player, it's said that this player breaks serve.

It is known that Alice won a games and Borys won b games during the match. It is unknown who served first and who won which games.

Find all values of k such that exactly k breaks could happen during the match between Alice and Borys in total.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.

Each of the next t lines describes one test case and contains two integers a and b (0 ≤ a, b ≤ 10^5; a + b > 0) — the number of games won by Alice and Borys, respectively.

It is guaranteed that the sum of a + b over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case print two lines.

In the first line, print a single integer m (1 ≤ m ≤ a + b + 1) — the number of values of k such that exactly k breaks could happen during the match.

In the second line, print m distinct integers k_1, k_2, …, k_m (0 ≤ k_1 < k_2 < … < k_m ≤ a + b) — the sought values of k in increasing order.

Example

Input


3
2 1
1 1
0 5


Output


4
0 1 2 3
2
0 2
2
2 3

Note

In the first test case, any number of breaks between 0 and 3 could happen during the match: 

  * Alice holds serve, Borys holds serve, Alice holds serve: 0 breaks; 
  * Borys holds serve, Alice holds serve, Alice breaks serve: 1 break; 
  * Borys breaks serve, Alice breaks serve, Alice holds serve: 2 breaks; 
  * Alice breaks serve, Borys breaks serve, Alice breaks serve: 3 breaks. 



In the second test case, the players could either both hold serves (0 breaks) or both break serves (2 breaks).

In the third test case, either 2 or 3 breaks could happen: 

  * Borys holds serve, Borys breaks serve, Borys holds serve, Borys breaks serve, Borys holds serve: 2 breaks; 
  * Borys breaks serve, Borys holds serve, Borys breaks serve, Borys holds serve, Borys breaks serve: 3 breaks. 
import java.io.*;
import java.util.*;

public class CF_1561b{
    public static final void main(String[] args){
        Kattio io= new Kattio();
        int t= io.getInt();
        while(t-->0){
            int a= io.getInt(), b= io.getInt(), extra= 0;
            int games= a+b;
            if(games%2==1){
                if(a>0){
                    a-= 1;
                    extra+= 1;
                }else{
                    b-= 1;
                    extra+= 1;
                }
            }
            // a+b is now even
            int halfAB= (a+b)/2;
            int minBreaks= Math.abs(a-halfAB), maxBreaks= a+b-minBreaks;

            List<String> ks= new ArrayList<>();
            for(int i=minBreaks; i<=maxBreaks; i+=2){
                ks.add(String.valueOf(i));
                if(extra!=0) ks.add(String.valueOf(i+1));
            }            
            io.println(ks.size());
            io.println(String.join(" ", ks));
        }
        io.close();
    }

    static void err(String format, Object... args){
        System.err.println(String.format(format, args));
    }    
    static class Kattio extends PrintWriter {
        private BufferedReader r;
        private String line, token;
        private StringTokenizer st;
    
        public Kattio(){this(System.in);}
        public Kattio(InputStream i){
        super(new BufferedOutputStream(System.out));
            r= new BufferedReader(new InputStreamReader(i));
        }
        public Kattio(InputStream i, OutputStream o){
        super(new BufferedOutputStream(o));
            r= new BufferedReader(new InputStreamReader(i));
        }
        public boolean isLineEnd(){
            return !st.hasMoreTokens();
        }
        public boolean hasMoreTokens(){
            return peekToken()!=null;
        }
        public int getInt(){
            return Integer.parseInt(nextToken());
        }
        public double getDouble(){ 
            return Double.parseDouble(nextToken());
        }
        public long getLong(){
            return Long.parseLong(nextToken());
        }
        public String getWord(){
            return nextToken();
        }
        private String peekToken(){
            if(token==null) try {
                while(st==null || !st.hasMoreTokens()) {
                    line= r.readLine();
                    if(line==null) return null;
                    st= new StringTokenizer(line);
                }
                token= st.nextToken();
            }catch(IOException e){}
            return token;
        }
        private String nextToken() {
            String ans= peekToken();
            token= null;
            return ans;
        }
    }    
}