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.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Objects;
import java.util.Scanner;
import java.util.stream.Collectors;

public class P1561B {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int cases = sc.nextInt();
        while (cases-- > 0) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            if (a < b) {
                solve(a, b);
            } else {
                solve(b, a);
            }
        }
    }

    private static void solve(int a, int b) {
        HashSet<Integer> ans = new HashSet<>();
        int total = a + b;
        if (total % 2 == 0) {
            int serve = total / 2;
            for (int holds = 0; holds <= a; holds++) {
                ans.add(serve - holds + (a - holds));
            }
        } else {
            int serveA = total / 2;
            int serveB = serveA + 1;
            for (int holds = 0; holds <= a; holds++) {
                ans.add(serveA - holds + (a - holds));
                ans.add(serveB - holds + (a - holds));
            }
        }
        ArrayList<Integer> list = new ArrayList<>(ans);
        List<Integer> sorted = list.stream().sorted().collect(Collectors.toList());
        System.out.println(sorted.size());
        String s = String.join(" ", sorted.stream().map(Objects::toString).collect(Collectors.toList()));
        System.out.println(s);
    }
}
