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;
  #pragma GCC target("avx2")
  #pragma GCC optimization("O3")
  #pragma GCC optimization("unroll-loops")
  constexpr long long pct(long long x) { return __builtin_popcount(x); }
  const long long DFS_WHITE = 0;
  const long long DFS_BLACK = 1;
  const long long INF = 1e18;
  const long long MAXN = 2e5 + 10;
  const long long MOD = 1e9 + 7;
  void setIO(string filename = "") {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    if (filename.size() == 0) return;
    freopen((filename + ".in").c_str(), "r", stdin);
    freopen((filename + ".out").c_str(), "w", stdout);
  }
  long long n, m;
  void solve() {
    cin >> n >> m;
    if (m > n) {
      swap(n, m);
    }
    if ((m + n) % 2 == 0) {
      long long tmp = (n + m) / 2;
      vector<long long> ans;
      for (long long i = n - tmp; i <= n - tmp + m * 2 and i <= n + m; i += 2) {
        ans.push_back(i);
      }
      cout << ans.size() << endl;
      for (auto a : ans) {
        cout << a << " ";
      }
    } else {
      vector<long long> ans;
      long long tmp = (n + m + 1) / 2;
      for (long long i = n - tmp; i <= n - tmp + m * 2 and i <= n + m; i += 2) {
        ans.push_back(i);
      }
      tmp = (n + m - 1) / 2;
      for (long long i = n - tmp; i <= n - tmp + 2 * m and i <= n + m; i += 2) {
        ans.push_back(i);
      }
      vector<long long> res;
      long long cur = 0;
      res.push_back(ans[cur]);
      for (long long i = 0; i < ans.size(); i++) {
        if (ans[i] != ans[cur]) {
          cur++;
          res.push_back(ans[i]);
        }
      }
      sort(ans.begin(), ans.end());
      cout << ans.size() << endl;
      for (long long i = 0; i < ans.size(); i++) {
        cout << ans[i] << " ";
      }
    }
  }
  signed main() {
    setIO();
    long long test = 1;
    cin >> test;
    while (test--) {
      solve();
      if (test) cout << endl;
    }
  }
  