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;
  const int OO = 1e9;
  const double EPS = 1e-9;
  template <class T>
  void _db(const char* dbStr, T e) {
    cout << dbStr << " = " << e << endl;
  }
  template <class T, class... L>
  void _db(const char* dbStr, T e, L... r) {
    while (*dbStr != ',') cout << *dbStr++;
    cout << " = " << e << ',';
    _db(dbStr + 1, r...);
  }
  template <class S, class T>
  ostream& operator<<(ostream& o, const map<S, T>& v) {
    o << "[";
    int i = 0;
    for (const pair<S, T>& pr : v)
      o << (!i++ ? "" : ", ") << "{" << pr.first << " : " << pr.second << "}";
    return o << "]";
  }
  template <template <class, class...> class S, class T, class... L>
  ostream& operator<<(ostream& o, const S<T, L...>& v) {
    o << "[";
    int i = 0;
    for (const auto& e : v) o << (!i++ ? "" : ", ") << e;
    return o << "]";
  }
  template <class S, class T>
  ostream& operator<<(ostream& o, const pair<S, T>& pr) {
    return o << "(" << pr.first << ", " << pr.second << ")";
  }
  ostream& operator<<(ostream& o, const string& s) {
    for (const char& c : s) o << c;
    return o;
  }
  template <class T>
  using V = vector<T>;
  template <class T>
  using VV = V<V<T>>;
  template <class T>
  using VVV = VV<V<T>>;
  using ll = long long;
  using pii = pair<int, int>;
  using vi = V<int>;
  using vii = V<pii>;
  using vvi = VV<int>;
  using mii = map<int, int>;
  using umii = unordered_map<int, int>;
  using si = set<int>;
  using usi = unordered_set<int>;
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  int main() {
    ios::sync_with_stdio(false);
    cout.precision(10);
    cin.tie(0);
    int T;
    cin >> T;
    for (int tc = 1; tc <= T; ++tc) {
      int a, b;
      cin >> a >> b;
      si ans;
      for (int I = 0; I <= 1; ++I) {
        int odd = (a + b + 1) / 2, even = (a + b) / 2;
        for (int x = 0; x <= min(odd, a); ++x) {
          int rem = a - x;
          if (rem > even) continue;
          ans.insert((odd - x) + rem);
        }
        swap(a, b);
      }
      cout << int(ans.size()) << '\n';
      for (auto& elem : ans) {
        cout << elem << ' ';
      }
      cout << '\n';
    }
    return 0;
  }
  