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 M = 1e9 + 7;
  void _print(vector<vector<long long>> v) {
    for (int i = 0; i < v.size(); ++i) {
      vector<long long> b = v[i];
      cerr << "[ ";
      for (int j = 0; j < b.size(); ++j) {
        cerr << v[i][j] << " ";
      }
      cerr << "]";
      cerr << endl;
    }
  }
  void _print(vector<long long> v) {
    cerr << "["
         << " ";
    for (long long i : v) {
      cerr << i << " ";
    }
    cerr << "]";
  }
  void _print(vector<bool> v) {
    cerr << "["
         << " ";
    for (long long i : v) {
      cerr << i << " ";
    }
    cerr << "]";
  }
  void _print(vector<pair<long long, long long>> v) {
    cerr << "[" << endl;
    for (auto i : v) {
      cerr << i.first << " " << i.second << endl;
    }
    cerr << "]";
  }
  void _print(map<long long, long long> mp) {
    cerr << "[" << endl;
    for (auto i : mp) {
      cerr << i.first << " " << i.second << endl;
    }
    cerr << "]";
  }
  void _print(unordered_map<long long, long long> mp) {
    cerr << "[" << endl;
    for (auto i : mp) {
      cerr << i.first << " " << i.second << endl;
    }
    cerr << "]";
  }
  void _print(set<long long> s) {
    cerr << "["
         << " ";
    for (auto i : s) {
      cerr << i << " ";
    }
    cerr << "]";
  }
  void _print(multiset<long long> s) {
    cerr << "["
         << " ";
    for (auto i : s) {
      cerr << i << " ";
    }
    cerr << "]";
  }
  void _print(unordered_set<long long> s) {
    cerr << "["
         << " ";
    for (auto i : s) {
      cerr << i << " ";
    }
    cerr << "]";
  }
  void _print(vector<int> v) {
    cerr << "["
         << " ";
    for (int i : v) {
      cerr << i << " ";
    }
    cerr << "]";
  }
  void _print(vector<string> v) {
    cerr << "["
         << " ";
    for (auto i : v) {
      cerr << i << endl;
    }
    cerr << "]";
  }
  void _print(pair<long long, long long> a) {
    cerr << a.first << " " << a.second;
  }
  void _print(int a) { cerr << a; }
  void _print(long long a) { cerr << a; }
  void _print(string a) { cerr << a; }
  void _print(bool a) { cerr << a; }
  void _print(char a) { cerr << a; }
  void _print(double a) { cerr << a; }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
  long long bpow(long long a, long long b, long long mod = M) {
    long long res = 1;
    while (b) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b >>= 1;
    }
    return res;
  }
  bool isSorted(vector<long long> &a) {
    long long f = 1;
    for (int i = 0; i < a.size(); ++i) {
      if (a[i] > a[i + 1]) {
        f = 0;
        break;
      }
    }
    return f;
  }
  void solve() {
    long long a, b;
    cin >> a >> b;
    double k = (double)(a + b) / 2;
    long long p = ceil(k);
    long long q = floor(k);
    vector<long long> ans;
    for (int i = 0; i <= p; ++i) {
      long long y = a - (p - i);
      if (y >= 0 and y <= q) ans.push_back(i + y);
    }
    for (int i = 0; i <= q; ++i) {
      long long y = a - (q - i);
      if (y >= 0 and y <= p) ans.push_back(i + y);
    }
    sort(ans.begin(), ans.end());
    vector<long long> ans1;
    ans1.push_back(ans[0]);
    for (int i = 1; i < ans.size(); ++i) {
      if (ans[i] != ans[i - 1]) ans1.push_back(ans[i]);
    }
    cout << ans1.size() << endl;
    for (int i = 0; i < ans1.size(); ++i) {
      cout << ans1[i] << " ";
    }
    cout << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  