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 bin_pow(long long base, long long deg, long long md) {
    if (!deg) return 1;
    if (deg % 2) return base * bin_pow(base, deg - 1, md) % md;
    long long res = bin_pow(base, deg / 2, md);
    return res * res % md;
  }
  template <typename T1, typename T2>
  bool cmin(T1& a, const T2& b) {
    if (a <= b) return false;
    a = b;
    return true;
  }
  template <typename T1, typename T2>
  bool cmax(T1& a, const T2& b) {
    if (a >= b) return false;
    a = b;
    return true;
  }
  mt19937 rnd(0);
  long long get_rand_int(long long mn, long long mx) {
    return rnd() % (mx - mn + 1) + mn;
  }
  const long long md = 1e9 + 7;
  long long inv(long long a) { return bin_pow(a, md - 2, md); }
  long long gcd(long long a, long long b) {
    if (!b) return a;
    return gcd(b, a % b);
  }
  long long get_bit(long long m, long long index) {
    return bool((m >> index) & 1ll);
  }
  long long get_fact(long long n) {
    if (n == 0) return 1;
    return n * get_fact(n - 1) % md;
  }
  long long get_c(long long n, long long k) {
    return get_fact(n) / get_fact(k) / get_fact(n - k);
  }
  void debug_assert(bool a) { assert(a); }
  set<long long> result;
  bool is_correct(long long num, long long max) { return num >= 0 && num <= max; }
  void solve(long long a, long long b) {
    long long first_play = (a + b + 1) / 2;
    long long first_get = a + b - first_play;
    for (long long x = 0; x <= first_get; ++x) {
      long long first_get_lost = first_get - x;
      long long first_play_win = a - x;
      long long first_play_lost = first_play - first_play_win;
      if (!is_correct(x, a) || !is_correct(first_play_win, a) ||
          !is_correct(first_get_lost, b) || !is_correct(first_play_lost, b))
        continue;
      long long res = x + first_play_lost;
      if (res == 4) res = 4;
      result.insert(x + first_play_lost);
    }
  }
  void run() {
    long long a, b;
    cin >> a >> b;
    solve(a, b);
    solve(b, a);
    cout << long long(result.size()) << '\n';
    for (auto u : result) {
      cout << u << ' ';
    }
    cout << '\n';
    result.clear();
  }
  signed main() {
    cin.tie(0);
    cout.tie(0);
    ios_base::sync_with_stdio(0);
    long long t = 1;
    cin >> t;
    for (long long i = 0; i < t; ++i) {
      run();
    }
  }
  