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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashSet;
import java.util.StringTokenizer;

public class B{
    public static void main(String[] args){
        FS sc = new FS();
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
        while(t-->0){
            int a = sc.nextInt();
            int b = sc.nextInt();
            int n = a+b;
            a = Math.min(a, b);
            b = Math.max(a, b);
            HashSet<Integer> set = new HashSet<>();
            for(int i=0; i<=Math.min(a, n/2); i++){
                // set.add(n/2-(a-i));
                set.add(n/2-i+(a-i));
                // set.add(n-(n/2-(a-i)));
                set.add(n-(n/2-i+(a-i)));
            }
            StringBuilder sb = new StringBuilder();
            sb.append(set.size()+"\n");
            for(int val : set){
                sb.append(val+" ");
            }
            System.out.println(sb);
        }
        pw.flush();
        pw.close();
    }

    static class FS{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
    
        String next(){
            while(!st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                } catch(Exception ignored){
                }
            }
            return st.nextToken();
        }
        
        int[] nextArray(int n){
            int[] a = new int[n];
            for(int i = 0; i < n; i++){
                a[i] = nextInt();
            }
            return a;
        }
        
        long[] nextLongArray(int n){
            long[] a = new long[n];
            for(int i = 0; i < n; i++){
                a[i] = nextLong();
            }
            return a;
        }
        
        int nextInt(){
            return Integer.parseInt(next());
        }
        
        long nextLong(){
            return Long.parseLong(next());
        }
        
        double nextDouble(){
            return Double.parseDouble(next());
        }
    }
}