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: /*
      Author : Akshat Jindal
      from NIT Jalandhar , Punjab , India
      JAI MATA DI
   */
  import java.util.*;
  
  
  import java.io.*;
  import java.math.*;
  import java.sql.Array;
  
  
  
  public class Main {
  	  static class FR{
  		  
  			BufferedReader br;
  			StringTokenizer st;
  			public FR() {
  				br = new BufferedReader(new InputStreamReader(System.in));
  			}
  			String next() 
  		    { 
  		        while (st == null || !st.hasMoreElements()) 
  		        { 
  		            try
  		            { 
  		                st = new StringTokenizer(br.readLine()); 
  		            } 
  		            catch (IOException  e) 
  		            { 
  		                e.printStackTrace(); 
  		            } 
  		        } 
  		        return st.nextToken(); 
  		    } 
  
  		    int nextInt() 
  		    { 
  		        return Integer.parseInt(next()); 
  		    } 
  
  		    long nextLong() 
  		    { 
  		        return Long.parseLong(next()); 
  		    } 
  
  		    double nextDouble() 
  		    { 
  		        return Double.parseDouble(next()); 
  		    } 
  
  		    String nextLine() 
  		    { 
  		        String str = ""; 
  		        try
  		        { 
  		            str = br.readLine(); 
  		        } 
  		        catch (IOException e) 
  		        { 
  		            e.printStackTrace(); 
  		        } 
  		        return str; 
  		    } 
  		}
  	 
  	  static long mod = (long)(1e9 + 7);
  	 
  	static void sort(long[] arr ) {
  		 ArrayList<Long> al = new ArrayList<>();
  		 for(long e:arr) al.add(e);
  		 Collections.sort(al);
  		 for(int i = 0 ; i<al.size(); i++) arr[i] = al.get(i);
  	 }
  	static void sort(int[] arr ) {
  		 ArrayList<Integer> al = new ArrayList<>();
  		 for(int e:arr) al.add(e);
  		 Collections.sort(al);
  		 for(int i = 0 ; i<al.size(); i++) arr[i] = al.get(i);
  	 }
  	static int UB(long[] arr , long find , int l , int r) {
  		while(l<=r) {
  			int m = (l+r)/2;
  			if(arr[m]<find) l = m+1;
  			else r = m-1;
  		}
  		return l;
  	}
  	static int LB(long[] arr , long find,int l ,int r ) {
  		while(l<=r) {
  			int m = (l+r)/2;
  			if(arr[m] > find) r = m-1;
  			else l = m+1;
  		}
  		return r;
  	}
  	static int UB(int[] arr , long find , int l , int r) {
  		while(l<=r) {
  			int m = (l+r)/2;
  			if(arr[m]<find) l = m+1;
  			else r = m-1;
  		}
  		return l;
  	}
  	static int LB(int[] arr , long find,int l ,int r ) {
  		while(l<=r) {
  			int m = (l+r)/2;
  			if(arr[m] > find) r = m-1;
  			else l = m+1;
  		}
  		return r;
  	}
  	 static long gcd(long a, long b)
  	  {      
  	     if (b == 0)
  	        return a;
  	     return gcd(b, a % b); 
  	  }
  	 static boolean[] prime(int num) {
  			boolean[] bool = new boolean[num];
  		     
  		      for (int i = 0; i< bool.length; i++) {
  		         bool[i] = true;
  		      }
  		      for (int i = 2; i< Math.sqrt(num); i++) {
  		         if(bool[i] == true) {
  		            for(int j = (i*i); j<num; j = j+i) {
  		               bool[j] = false;
  		            }
  		         }
  		      }
  		      if(num >= 0) {
  		    	  bool[0] = false;
  		    	  bool[1] = false;
  		      }
  		      
  		      return bool;
  		}
  	 static void sort(char[] arr) {
  		 ArrayList<Character> al = new ArrayList<Character>();
  		 for(char cc:arr) al.add(cc);
  		 Collections.sort(al);
  		 for(int i = 0 ;i<arr.length ;i++) arr[i] = al.get(i);
  		 
  	 }
  	 static long[][] ncr(int n, int k)
  	    {
  	        long C[][] = new long[n + 1][k + 1];
  	        int i, j;
  	 
  	        // Calculate  value of Binomial
  	        // Coefficient in bottom up manner
  	        for (i = 0; i <= n; i++) {
  	            for (j = 0; j <= Math.min(i, k); j++) {
  	                // Base Cases
  	                if (j == 0 || j == i)
  	                    C[i][j] = 1;
  	 
  	                // Calculate value using
  	                // previously stored values
  	                else
  	                    C[i][j] = (C[i - 1][j - 1] + C[i - 1][j])%mod;
  	            }
  	        }
  	 
  	        return C;
  	    }
  	 static long modInverse(long a, long m)
  	    {
  	        long g = gcd(a, m);
  	       
  	          return   power(a, m - 2, m);
  	        
  	    }
  	   
  	      static long power(long x, long y, long m)
  	    {
  	        if (y == 0)
  	            return 1;
  	        long p = power(x, y / 2, m) % m;
  	        p = (int)((p * (long)p) % m);
  	        if (y % 2 == 0)
  	            return p;
  	        else
  	            return (int)((x * (long)p) % m);
  	    }
  	      static class DisjointUnionSets {
  			    int[] rank, parent;
  			    int n;
  			  
  			    public DisjointUnionSets(int n)
  			    {
  			        rank = new int[n];
  			        parent = new int[n];
  			        this.n = n;
  			        makeSet();
  			    }
  			  
  			    void makeSet()
  			    {
  			        for (int i = 0; i < n; i++) {
  			          
  			            parent[i] = i;
  			        }
  			    }
  			  
  			    int find(int x)
  			    {
  			        if (parent[x] != x) {
  			        
  			            parent[x] = find(parent[x]);
  			  
  			        }
  			  
  			        return parent[x];
  			    }
  			  
  			    void union(int x, int y)
  			    {
  			        int xRoot = find(x), yRoot = find(y);
  			  
  			        if (xRoot == yRoot)
  			            return;
  			  
  			        if (rank[xRoot] < rank[yRoot])
  			  
  			            parent[xRoot] = yRoot;
  			  
  			        else if (rank[yRoot] < rank[xRoot])
  			  
  			            parent[yRoot] = xRoot;
  			  
  			        else
  			        {
  			            parent[yRoot] = xRoot;
  			  
  			            rank[xRoot] = rank[xRoot] + 1;
  			        }
  			    }
  			}
  
  /* ***************************************************************************************************************************************************/	 
  	 
  	      static FR sc = new FR();
  	 static StringBuilder sb = new StringBuilder();
  	 public static void main(String args[]) {
  		
  		 int tc = 1;
  		  tc = sc.nextInt();
  
  		 while(tc-->0) {
  			 TEST_CASE();
  			 
  		 }
  		 
  	    System.out.println(sb);
  	 }
  	 
  	 static void TEST_CASE() {
  		 TreeSet<Integer> ts = new TreeSet<>();
  		 int a = sc.nextInt() , b = sc.nextInt();
  		 
  		 // a extra
  		 if(a >= b) {
  			 int add = (a-b)/2 + (a-b)%2;
  			 int rem = 2*b;
  			 for(int i = 0 ; i<=2*b ;i+=2 ) ts.add(i+add);
  		 }
  		 if(b>=a) {
  			 int add = (b-a)/2;
  			 for(int i =0 ; i<=2*a;i+=2) ts.add(i+add);
  		 }
  		 
  		 // b extra
  		 if(b>=a) {
  			 int add = (b-a)/2 + (b-a)%2;
  			 for(int i =0 ; i<=2*a;i+=2) ts.add(i+add);
  		 }
  		 if(a >= b) {
  			 int add = (a-b)/2;
  			 int rem = 2*b;
  			 for(int i = 0 ; i<=2*b ;i+=2 ) ts.add(i+add);
  		 }
  		 sb.append(ts.size()+"\n");
  		 while(!ts.isEmpty()) {
  			 int val = ts.first();
  			 ts.remove(val);
  			 sb.append(val+" ");
  		 }
  		 sb.append("\n");
  	 }
  	 
  }
  
  
  
  
  
  
  
  