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: import java.io.*;
  import java.util.ArrayList;
  import java.util.Collections;
  import java.util.StringTokenizer;
  
  public class Charmed {
  
      //Fast Input
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              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;
          }
      }
  
      //Printing an array
      static void print(long[][] a, int r, int c) {
          PrintWriter pw = new PrintWriter(System.out);
          for (int i = 0; i < r; i++) {
              for (int j = 0; j < c; j++) {
                  pw.print(a[i][j] + " ");
              }
              pw.println();
          }
          pw.flush();
      }
  
      //Used to generate random test cases
      //Must be changed according to input format
      static void testCaseGenerator(int t, int lowN, int highN, long lowA, long highA) throws FileNotFoundException {
          File input = new File("C:\\Users\\debar\\OneDrive\\Documents\\input.txt");
          PrintWriter pw = new PrintWriter(input);
          pw.println(t);
          for (int i = 0; i < t; i++) {
              int n = (int) (Math.random() * (highN - lowN)) + lowN;
              pw.println(n);
              for (int j = 0; j < n; j++) {
                  long a = (long) (Math.random() * (highA - lowA)) + lowA;
                  pw.print(a + " ");
              }
              pw.println();
          }
          pw.flush();
      }
  
      //Find factorial of numbers from 1 to n
      //Check the size of factorial array
      static long[] factorial(int n, long mod) {
          long[] factorial = new long[n];
          factorial[0] = 1;
          for (int i = 1; i <= n; i++) {
              factorial[i] = (factorial[i - 1] * i) % mod;
          }
          return factorial;
      }
  
      //Find factorial inverse from 1 to n
      //Check the size of inverse_factorial array
      static long[] factorialInverse(int n, long mod) {
          long[] inverse_factorial = modularInverse(n, mod);
          for (int i = 2; i <= n; i++) {
              inverse_factorial[i] = (inverse_factorial[i] * inverse_factorial[i - 1]) % mod;
          }
          return inverse_factorial;
      }
  
      //Inverse exist when two number are relatively prime
      //Based on Fermat little theorem (a^mod-1)%mod=1 when mod is prime
      //a * a^(mod - 2) % mod = 1
      //a^(mod - 2) % mod =a^-1
      //When mod != prime use Extended euclidean theorem to calculate x and y for equation a*x + mod*y = gcd(a,b)=1
      //(a*x + mod*y = gcd(a,b)=1) % mod = a*x %mod =1
      //https://www.geeksforgeeks.org/find-initial-integral-solution-of-linear-diophantine-equation-if-finite-solution-exists/?ref=rp
      //Used for only one number
      static long modularInverse(long a, long mod) {
          return binaryExponentiation(a, mod - 2, mod);
      }
  
      //Modular inverse for number from 1 to n - 1
      //Time complexity O(n) for n numbers
      //Proof https://cp-algorithms.com/algebra/module-inverse.html
      static long[] modularInverse(int n, long mod) {
          long[] inv = new long[n];
          inv[1] = 1;
          for (int i = 2; i < n; i++) {
              inv[i] = mod - (mod / i) * inv[(int) (mod % i)] % mod;
          }
          return inv;
      }
  
      //GCD calculation of a and b with complexity O(log(Max(a,b))
      static long gcd(long a, long b) {
          long r;
          while (b != 0) {
              r = a % b;
              a = b;
              b = r;
          }
          return a;
      }
  
      //Find prime number upto n with complexity O(n*log(log(n)))
      //Travel through the multiple of prime number and set it to false
      static boolean[] sieveOfEratosthenes(int n) {
  
          boolean[] prime = new boolean[n + 1];
          for (int i = 0; i <= n; i++)
              prime[i] = true;
  
          for (int p = 2; p * p <= n; p++) {
              if (prime[p]) {
                  for (int i = p * p; i <= n; i += p)
                      prime[i] = false;
              }
          }
  
          prime[0] = prime[1] = false;
          return prime;
      }
  
      //Find Prefix sum in O(n)
      static long[] PrefixSum(long[] arr, int n) {
  
          long[] prefixSum = new long[n];
  
          prefixSum[0] = arr[0];
          for (int i = 1; i < n; ++i)
              prefixSum[i] = prefixSum[i - 1] + arr[i];
          return prefixSum;
      }
  
      //Find a^n in O(log n)
      //If n=11 then n in binary is equal to 1011
      //a^n=a^(8+2+1) that is multiply with result when bit is set.
      //https://youtu.be/L-Wzglnm4dM
      static long binaryExponentiation(long a, long n, long mod) {
          long result = 1L;
          n = n % (mod - 1);
          //n = (mod - 1) * floor(n / (mod - 1)) + n % (mod - 1)
          //a^((mod - 1) * floor(n / (mod - 1))) % mod = 1 so a^n % mod =a^(n % (mod - 1)) % mod when mod is prime
          //Using fermat little theorem a^(p-1) % p = 1 when p is prime
          while (n > 0) {
              if ((n & 1) == 1) {
                  result = (result * a) % mod;
              }
              a = (a * a) % mod;
              n /= 2;
          }
          return result;
      }
  
      //Matrix multiplication
      static long[][] matrixMultiply(long[][] a, long[][] b, int row, int col, int mid, long mod) {
          long[][] c = new long[row][col];
          for (int i = 0; i < row; i++) {
              for (int j = 0; j < col; j++) {
                  c[i][j] = 0;
                  for (int k = 0; k < mid; k++) {
                      c[i][j] = (c[i][j] + (a[i][k] * b[k][j]) % mod) % mod;
                  }
              }
          }
          return c;
      }
  
      //Matrix Exponentiation
      static long[][] matrixExponentiation(long[][] a, long n, int size, long mod) {
          if (n == 0) {
              long[][] temp = new long[size][size];
              for (int i = 0; i < size; i++) {
                  temp[i][i] = 1;
              }
              return temp;//Returns identity matrix
          } else {
              long[][] temp = matrixExponentiation(a, n / 2, size, mod);
              if (n % 2 == 0) {
                  return matrixMultiply(temp, temp, size, size, size, mod);
              } else {
                  return matrixMultiply(matrixMultiply(temp, temp, size, size, size, mod), a, size, size, size, mod);
              }
          }
      }
  
      //Solution starts here
      public static void main(String[] args) {
  
          ArrayList<Integer> arr=new ArrayList<>();
  
          //Input output class reference
          FastReader sc = new FastReader();
          PrintWriter pw = new PrintWriter(System.out);
  
          //Number of test case
          int t = sc.nextInt();
  
          //For each test case
          for (int i = 0; i < t; i++) {
              int a=sc.nextInt();
              int b=sc.nextInt();
              int n=a+b;
              if((n&1)==1)
              {
                  int b1=Math.abs(a-b)/2;
                  int b2=b1+1;
                  int min1=Math.min(a,b);
                  int min=min1;
                  for(int j=b1;j<=n && min>=0;j+=2)
                  {
                      arr.add(j);
                      min--;
                  }
                  min=min1;
                  for(int j=b2;j<=n && min>=0;j+=2)
                  {
                      arr.add(j);
                      min--;
                  }
                  Collections.sort(arr);
  
                  pw.println(arr.size());
                  for(int j:arr)
                  {
                      pw.print(j+" ");
                  }
              }
              else
              {
                  int b1=Math.abs(a-b)/2;
                  int min=Math.min(a,b);
                  for(int j=b1;j<=n && min>=0;j+=2)
                  {
                      arr.add(j);
                      min--;
                  }
  
                  pw.println(arr.size());
                  for(int j:arr)
                  {
                      pw.print(j+" ");
                  }
              }
              pw.println();
              arr.clear();
          }
          pw.flush();
      }
      //Solution ends here
  }