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 mod = 1e5 + 5;
  int power(int x, int n) {
    int res = 1;
    while (n) {
      if (n & 1) res = 1LL * res * x % mod;
      x = 1LL * x * x % mod;
      n >>= 1;
    }
    return res;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      int a, b;
      cin >> a >> b;
      set<int> ways;
      int oddposs, evenposs;
      oddposs = (a + b) / 2;
      evenposs = (a + b) - oddposs;
      for (int i = 0; i < oddposs + 1; i++) {
        int ae, ao, be, bo;
        ao = i;
        ae = a - i;
        bo = oddposs - ao;
        be = evenposs - ae;
        if ((ae) < 0 || bo < 0 || be < 0) continue;
        ways.insert((ao + be));
      }
      for (int i = 0; i < evenposs + 1; i++) {
        int ae, ao, bo, be;
        ae = i;
        ao = a - i;
        bo = oddposs - ao;
        be = evenposs - ae;
        if (ao < 0 || (ae) < 0 || bo < 0 || be < 0) continue;
        ways.insert((ae + bo));
      }
      cout << ways.size() << endl;
      for (auto node : ways) cout << node << " ";
      cout << endl;
    }
    return 0;
  }
  