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: #include <bits/stdc++.h>
  using namespace std;
  long long int mul(long long int a, long long int b) {
    return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
  }
  long long int add(long long int a, long long int b) {
    return (a % 1000000007 + b % 1000000007) % 1000000007;
  }
  long long int gcd(long long int a, long long int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long int extgcd(long long int a, long long int b, long long int& x,
                       long long int& y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    long long int x1, y1;
    long long int d = extgcd(b, a % b, x1, y1);
    x = y1;
    y = x1 - y1 * (a / b);
    return d;
  }
  bool isPrime(long long int n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (long long int i = 5; i * i <= n; i = i + 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  bool isPowerOfTwo(long long int n) {
    if (n == 0) return false;
    return (ceil(log2(n)) == floor(log2(n)));
  }
  vector<long long int> factors(long long int n) {
    vector<long long int> f;
    for (long long int x = 2; x * x <= n; x++) {
      while (n % x == 0) {
        f.push_back(x);
        n /= x;
      }
    }
    if (n > 1) f.push_back(n);
    return f;
  }
  long long int bin_exp_mod(long long int a, long long int n) {
    long long int res = 1;
    a %= 1000000007;
    if (a == 0) return 0;
    while (n > 0) {
      if (n & 1) res = ((res % 1000000007) * (a % 1000000007)) % 1000000007;
      n = n >> 1;
      a = ((a % 1000000007) * (a % 1000000007)) % 1000000007;
    }
    res %= 1000000007;
    return res;
  }
  void solve() {
    long long int a, b;
    cin >> a >> b;
    long long int n = a + b;
    set<long long int> st;
    if ((a + b) % 2 == 0) {
      long long int x = (a + b) / 2;
      long long int y = (a + b) / 2;
      long long int d = abs(a - b) / 2;
      for (long long int i = d; i < a + b - d + 1; i++) {
        st.insert(i);
        i++;
      }
    } else {
      long long int x = (a + b) / 2 + 1;
      long long int y = (a + b) / 2;
      long long int d = abs(a - b) / 2;
      for (long long int i = d; i < a + b - d + 1; i++) {
        st.insert(i);
      }
    }
    cout << st.size() << "\n";
    for (auto x : st) cout << x << " ";
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    long long int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  