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;
  constexpr long long MOD = 998244353;
  constexpr long long mod = 1e9 + 7;
  const double eps = 1e-7;
  int dx[] = {0, 0, 1, -1};
  int dy[] = {1, -1, 0, 0};
  const long long N = 3e6 + 5;
  vector<long long> adj[N];
  long long power(long long n, long long p) {
    if (p == 0) return 1;
    if (p == 1) return n;
    if (p % 2)
      return power(n, p - 1) * n;
    else {
      long long x = power(n, p / 2);
      return x * x;
    }
  }
  long long modpow(long long a, long long b, long long m) {
    long long ans = 1;
    while (b) {
      if (b & 1) ans = (ans * a) % m;
      b /= 2;
      a = (a * a) % m;
    }
    return ans;
  }
  long long nsum(long long num) { return (num * (num + 1)) / 2; }
  void edge(long long u, long long v) {
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  void solve() {
    long long n, a, b;
    string s;
    cin >> a >> b;
    long long all = a + b;
    set<long long> st;
    long long aw = 0, ah = 0, bw = 0, bh = 0;
    for (long long i = 0; i <= all / 2; i++) {
      if (i > a) break;
      aw = i;
      ah = a - aw;
      bh = all / 2 - aw;
      bw = b - bh;
      if (ah < 0 || bh < 0 || aw < 0 || bw < 0) continue;
      if ((ah + aw) == a && (bh + bw) == b) {
        st.insert(aw + bw);
      }
    }
    for (long long i = 0; i <= (all + 1) / 2; i++) {
      if (i > a) break;
      aw = i;
      ah = a - aw;
      bh = (all + 1) / 2 - aw;
      bw = b - bh;
      if (ah < 0 || bh < 0 || aw < 0 || bw < 0) continue;
      if ((ah + aw) == a && (bh + bw) == b) {
        st.insert(aw + bw);
      }
    }
    cout << st.size() << "\n";
    for (auto it : st) cout << it << ' ';
    cout << "\n";
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long Test = 1;
    cin >> Test;
    while (Test--) {
      solve();
    }
    return 0;
  }
  