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.*;
public class Test2 {
	static Set<Integer> list;
	static long dp[][][];
	static long idx;
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			int a=sc.nextInt();
			int b=sc.nextInt();
			list=new HashSet<>();
			
			dp=new long[a+1][b+1][2];
			for(int i=0;i<a+1;i++)
				for(int j=0;j<b+1;j++)
					Arrays.fill(dp[i][j], -1);
			idx=0;
			
			util(0,0,true,a,b,0);
			util(0,0,false,a,b,0);
			System.out.println(list.size());
			ArrayList<Integer> ret=new ArrayList<>(list);
			Collections.sort(ret);
			for(int x:ret)
				System.out.print(x+" ");
			System.out.println();
		}
	}
	
	private static void util(int a,int b,boolean chance,int A,int B,int ans) {
		if(a+b==A+B) {
			list.add(ans);
			return ;
		}
		
		int x=1;
		if(!chance)
			x=0;
		if(dp[a][b][x]!=-1)
			return;
		
		dp[a][b][x]=idx++;
		
		//chance=true->A;false->B
		boolean next=!chance;
		if(a!=A && b!=B) {
			if(chance) {
				util(a+1,b,next,A,B,ans);
				util(a,b+1,next,A,B,ans+1);
			}else {
				util(a+1,b,next,A,B,ans+1);
				util(a,b+1,next,A,B,ans);
			}
		}else if(a!=A && b==B) {
			if(chance) {
				util(a+1,b,next,A,B,ans);
			}else {
				util(a+1,b,next,A,B,ans+1);
			}
		}else {
			if(chance) {
				util(a,b+1,next,A,B,ans+1);
			}else {
				util(a,b+1,next,A,B,ans);
			}
		}
	}

}
