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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class radix {
	static int[] a;

	/*
	 * boolean DHD() { for (int i = 0; i < a.length - 1; i++) { if (a[i] > a[i + 1])
	 * { return false; } } return true; }
	 */
	void solve() throws IOException {
		int a, b;
		int A; // Максимальное количество брейков Алисы
		int B; // Максимальное количество брейков Бориса
		int aMin;
		int bMin;
		int N = in.nextInt();
		for (int i = 0; i < N; i++) {
			a = in.nextInt();
			b = in.nextInt();
			A = (int) Math.min(Math.round((a + b) / 2.0), a);// 1
			B = (int) Math.min(Math.round((a + b) / 2.0), b);// 1
			aMin = (int) Math.min(a - Math.min(a, (a + b) / 2), a - Math.min(a, Math.round((a + b) / 2.0)));
			bMin = (int) Math.min(b - Math.min(b, (a + b) / 2), b - Math.min(b, Math.round((a + b) / 2.0)));
			if ((a + b) % 2 != 0) {
				out.println(A + B - aMin - bMin + 1);
				for (int j = aMin + bMin; j <= A + B; j++) {
					out.print(j + " ");
				}
				out.println();
				// for
			} else {
				out.println(A + B - aMin - bMin);
				for (int j = aMin + bMin; j <= A + B; j = j + 2) {
					out.print(j + " ");
				}
				out.println();
			}

		}
	}
	/*
	 * 3 2 1 1 1 0 5
	 */

	FastScanner in;
	static PrintWriter out;

	public void run() {
		try {
			in = new FastScanner();
			out = new PrintWriter(System.out);

			solve();

		} catch (Throwable e) {
			e.printStackTrace();
		} finally {
			out.close();
		}
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		boolean eof;

		FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		boolean hasNext() {
			if (st == null || !st.hasMoreTokens()) {
				return false;
			}
			return true;
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {
					eof = true;
					return null;
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			String s = next();
			return eof ? null : Integer.parseInt(s);
		}
	}

	public static void main(String[] arg) {
		new radix().run();
	}

}
