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 long double PI = 3.14159265359;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long randint(long long a, long long b) {
    return uniform_int_distribution<long long>(a, b)(rng);
  }
  template <typename... Args>
  void logger(string vars, Args&&... values) {
    cout << vars << " = ";
    string delim = "";
    (..., (cout << delim << values, delim = ", "));
    cout << endl;
  }
  void print(vector<int>& v) {
    cout << "[";
    for (int i = 0; i < v.size(); i++) {
      cout << v[i];
      if (i + 1 != v.size()) cout << ", ";
    }
    cout << "]" << endl;
  }
  void print(pair<int, int>& p) {
    cout << "{" << p.first << ", " << p.second << "}" << endl;
  }
  vector<int> junta(pair<int, int> a, pair<int, int> b) {
    int mini = 0;
    int maxi = 1;
    if (a.first > b.first) {
      mini = a.first - b.first;
      maxi += a.second;
    } else {
      mini = a.second - b.second;
      maxi += a.first;
    }
    vector<int> v;
    int cont = 0;
    for (int i = mini; i <= a.first + a.second && (cont++) < maxi; i += 2) {
      v.push_back(i);
    }
    return v;
  }
  signed main() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) {
      int a, b;
      cin >> a >> b;
      int n = (a + b);
      pair<int, int> t1 = {n / 2, (n + 1) / 2};
      pair<int, int> t2 = {(n + 1) / 2, n / 2};
      auto v1 = junta({a, b}, t1);
      auto v2 = junta({a, b}, t2);
      for (auto k : v2) {
        v1.push_back(k);
      }
      sort(v1.begin(), v1.end());
      auto un = unique(v1.begin(), v1.end());
      int cont = 0;
      for (auto i = v1.begin(); i != un; i++) {
        cont++;
      }
      cout << cont << '\n';
      for (auto i = v1.begin(); i != un; i++) {
        cout << *i << ' ';
      }
      cout << '\n';
    }
  }
  