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;
  template <typename T>
  istream &operator>>(istream &in, vector<T> &v) {
    for (auto &i : v) in >> i;
    return in;
  }
  template <typename A, typename B>
  istream &operator>>(istream &in, pair<A, B> &a) {
    in >> a.first >> a.second;
    return in;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long rand(long long l, long long r) {
    uniform_int_distribution<long long> uid(l, r);
    return uid(rng);
  }
  const string no = "No";
  const string yes = "Yes";
  void solve() {
    long long a, b;
    cin >> a >> b;
    long long rounds = a + b;
    long long matcha = (rounds + 1) / 2, matchb = rounds - matcha;
    set<long long> s;
    for (long long i = 0; i <= matchb; i++) {
      long long pointsa = matcha + i;
      long long pointsb = matchb - i;
      long long breaksfroma = pointsa - a;
      pointsa -= breaksfroma;
      pointsb += breaksfroma;
      if (pointsa < i) {
        long long takeback = i - pointsa;
        breaksfroma -= takeback;
        pointsa += takeback;
        pointsb += takeback;
      };
      ;
      if (b == pointsb && a == pointsa && breaksfroma >= 0) {
        s.insert(i + breaksfroma);
      }
    };
    matcha = rounds / 2, matchb = rounds - matcha;
    for (long long i = 0; i <= matcha; i++) {
      long long pointsb = matchb + i;
      long long pointsa = matcha - i;
      long long breaksfromb = pointsb - b;
      pointsb -= breaksfromb;
      pointsa += breaksfromb;
      if (pointsb < i) {
        long long takeback = i - pointsb;
        breaksfromb -= takeback;
        pointsb += takeback;
        pointsa -= takeback;
      };
      ;
      if (b == pointsb && a == pointsa && breaksfromb >= 0) {
        s.insert(i + breaksfromb);
      }
    };
    cout << (long long)(s).size() << '\n';
    for (long long i : s) {
      cout << i << ' ';
    }
    cout << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    cin >> t;
    for (long long tc = 1; tc <= t; tc++) {
      ;
      solve();
    }
    return 0;
  }
  