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 = 1e9 + 7;
  long long fxp(long long a, long long b) {
    assert(b >= 0);
    a %= mod;
    long long ans = 1;
    while (b > 0) {
      if (b % 2) ans = (ans * a) % mod;
      a = (a * a) % mod;
      b /= 2;
    }
    return ans % mod;
  }
  inline long long minv(long long x) { return fxp(x, mod - 2); }
  void c_p_c() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  const int mXn = 2e5 + 4;
  const long long inf = 1e18;
  vector<int> ans;
  int ck = 0;
  void brute(int lev, int arem, int brem) {
    if (arem + brem == 0) {
      ans.push_back(ck);
      return;
    }
    if (arem) {
      if (lev == 1) ck++;
      brute(lev ^ 1, arem - 1, brem);
      if (lev == 1) ck--;
    }
    if (brem) {
      if (lev == 0) ck++;
      brute(lev ^ 1, arem, brem - 1);
      if (lev == 0) ck--;
    }
  }
  signed main() {
    c_p_c();
    int T;
    cin >> T;
    for (long long tc = 1; tc <= T; tc++) {
      int a, b;
      cin >> a >> b;
      int s = (a + b);
      pair<int, int> x[2];
      x[0] = {s / 2, s - s / 2};
      x[1] = {s - s / 2, s / 2};
      int mn = max(0, a - x[0].first) + max(0, b - x[0].second);
      mn = min(mn, max(0, a - x[1].first) + max(0, b - x[1].second));
      int mx = min(a, x[0].second) + min(b, x[0].first);
      mx = max(mx, min(a, x[1].second) + min(b, x[1].first));
      vector<int> bb;
      if (s % 2) {
        for (long long i = mn; i < mx + 1; i++) bb.push_back(i);
      } else {
        for (int i = mn; i <= mx; i += 2) bb.push_back(i);
      }
      cout << bb.size() << '\n';
      for (auto x : bb) cout << x << ' ';
      cout << '\n';
    }
  }
  