description: 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. 
solution: //package codeforces;
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  import java.io.PrintWriter;
  public class codeforces {
  	public static void main(String[] args) {
  		FastScanner s=new FastScanner();
  		PrintWriter out=new PrintWriter(System.out);
  		int t=s.nextInt();
  		for(int tt=0;tt<t;tt++) {
  			int a=s.nextInt(),b=s.nextInt();
              if ((a + b) % 2 == 0) {
                  int each = (a + b) / 2;
                  // Alice can break serve
                  int max = Math.min(a, b) + each;
                  int min = each - Math.min(a, b);
                  out.println((max / 2 - min / 2 + 1));
                  for (int i = min; i <= max; i += 2) out.print(i + " ");
                  out.println();
              } else {
                  int each = (a + b) / 2;
                  // Alice can break serve
                  int max = Math.min(a, b) + each + 1;
                  int min = each - Math.min(a, b);
                  out.println((max - min + 1));
                  for (int i = min; i <= max; i++) out.print(i + " ");
              }
  		}
  		out.close();
  	}
  	static boolean isPrime(int n){
          if (n <= 1)
              return false;
          else if (n == 2)
              return true;
          else if (n % 2 == 0)
              return false;
          for (int i = 3; i <= Math.sqrt(n); i += 2){
              if (n % i == 0)
                  return false;
          }
          return true;
      }
      static void sort(long [] a) {
  		ArrayList<Long> l=new ArrayList<>();
  		for (long i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
      static void sort(int [] a) {
  		ArrayList<Integer> l=new ArrayList<>();
  		for (int i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
      static int gcd(int a, int b){
        if (b == 0)
          return a;
        return gcd(b, a % b);
      }
      static void sortcol(int a[][],int c) {
  		Arrays.sort(a, (x, y) -> {
  			if(x[c]>y[c]) {
  				return 1;
  			}else {
  				return -1;
  			}
  		});
  	}
      public static void printb(boolean ans) {
      	if(ans) {
      		System.out.println("Yes");
      	}else {
      		System.out.println("No");
      	}
      }
      static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		boolean nextBoolean() {
  			return Boolean.parseBoolean(next());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		long[] readArrayLong(int n) {
  			long[] a=new long[n];
  			for (int i=0; i<n; i++) a[i]=nextLong();
  			return a;
  		}
  	}
  	
  }
  class Pair implements Comparable<Pair>{
  	int a , b; 
  	Pair(int x , int y){
  		a=x;
  		b=y;
  	}
  	public int compareTo(Pair o) {
  		if(a!=o.a) {
  			return  a - o.a;
  		}
  		return  b - o.b;
  	}
  }
  