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 Contest1561B
{
	static class InputReader {
		BufferedReader reader;
		StringTokenizer tokenizer;
		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}
		String next() { // reads in the next string
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt() { // reads in the next int
			return Integer.parseInt(next());
		}
		public long nextLong() { // reads in the next long
			return Long.parseLong(next());
		}
		public double nextDouble() { // reads in the next double
			return Double.parseDouble(next());
			}
		}
	static InputReader r = new InputReader(System.in);
	static PrintWriter pw = new PrintWriter(System.out);
	public static void main(String[] args)
	{
		//LOOK FOR INT OVERFLOW
		int t = r.nextInt();
		while (t > 0)
		{
			t--;
			int a = r.nextInt(); int b = r.nextInt();
			if ((a+b)%2 == 0)
			{
				int n = (a+b)/2;
				int max = Math.min(a, n) + Math.min(b,n);
				int min = Math.abs(b-a)/2;
				int k = max - min + 1;
				pw.println(k);
				for (int i = 0; i < k - 1; i ++)
				{
					pw.print(min + i + " ");
				}
				pw.println(min + k - 1);
			}
			else
			{
				int c = Math.max(a, b); int e = Math.min(a, b);  a = c; b = e;
				int n = (a+b)/2;
				int max = Math.min(a, n+1) + Math.min(b, n);
				int min = n - b;
				int k = max - min  +1;
				pw.println(k);
				for (int i= 0; i < k - 1; i ++)
				{
					pw.print(min + i + " ");
				}
				pw.println(min + k - 1);
			}
		}
		
		pw.close();
	}
}