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;
  vector<string> vec_splitter(string s) {
    s += ',';
    vector<string> res;
    while (!s.empty()) {
      res.push_back(s.substr(0, s.find(',')));
      s = s.substr(s.find(',') + 1);
    }
    return res;
  }
  void debug_out(vector<string> __attribute__((unused)) args,
                 __attribute__((unused)) int idx,
                 __attribute__((unused)) int LINE_NUM) {
    cerr << '\n';
  }
  template <typename Head, typename... Tail>
  void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
    if (idx > 0)
      cerr << ", ";
    else
      cerr << "Line(" << LINE_NUM << ") ";
    stringstream ss;
    ss << H;
    cerr << args[idx] << " = " << ss.str();
    debug_out(args, idx + 1, LINE_NUM, T...);
  }
  void run() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
  }
  const int N = 1e5 + 5, mod = 1e9 + 7, mod2 = 998244353, MAXN = 100000 + 29;
  void files() {}
  int arr[N], n;
  bool go(bool f) {
    if (f)
      for (int i = 0; i < n - 1; i += 2) {
        if (arr[i] > arr[i + 1]) swap(arr[i], arr[i + 1]);
      }
    else
      for (int i = 1; i < n - 1; i += 2) {
        if (arr[i] > arr[i + 1]) swap(arr[i], arr[i + 1]);
      }
    bool ret = true;
    for (int i = 0; i < n; ++i) {
      ret &= arr[i] == (i + 1);
    }
    return ret;
  }
  int a, b;
  bool solve(int i, int j, int k, bool t) {
    if (i < 0 || j < 0) return 0;
    if (i == 0 && j == 0) {
      return k == 0;
    }
    if (k < 0) return 0;
    if (t) {
      return solve(i - 1, j, k, t ^ 1) || solve(i, j - 1, k - 1, t ^ 1);
    } else
      return solve(i, j - 1, k, t ^ 1) || solve(i - 1, j, k - 1, t ^ 1);
  }
  void solve() {
    cin >> a >> b;
    int diff = abs(a - b);
    vector<int> v;
    for (int i = diff / 2; i <= a + b - diff / 2;) {
      v.push_back(i);
      i += 1 + (diff % 2 == 0);
    }
    cout << v.size() << '\n';
    for (int i = 0; i < v.size(); ++i) {
      cout << v[i] << ' ';
    }
    cout << '\n';
  }
  int main() {
    files();
    int t = 1;
    cin >> t;
    while (t--) solve();
  }
  