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 long double PI = 3.141592653589793;
  const long long mod = 1e9 + 7;
  int CASE = 1;
  const int mxn = 1e5 + 1;
  const long long infll = 1e18;
  const int infi = 1e9;
  bool prime(long long n) {
    if (n <= 1) return false;
    if (n == 2 or n == 3) return true;
    if (n % 2 == 0 or n % 3 == 0) return false;
    for (long long i = 5; i * i <= n; i += 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  long long __gcd(long long a, long long b) { return !b ? a : __gcd(b, a % b); }
  long long power(long long a, long long b) {
    long long x = a, res = 1, p = b;
    while (p > 0) {
      if (p & 1) res *= x;
      x *= x;
      p >>= 1;
    }
    return res;
  }
  void solve() {
    int a, b;
    cin >> a >> b;
    set<int> s;
    if (a > b) swap(a, b);
    int n = (a + b) / 2;
    int m = a + b - n;
    for (int i = 0; i <= a; i++) {
      if (i <= n) {
        s.insert(i + b - (n - i));
      } else {
        s.insert(i + m - (i - n));
      }
      if (i <= m) {
        s.insert(i + b - (m - i));
      } else {
        s.insert(i + n - (i - m));
      }
    }
    cout << s.size() << '\n';
    for (int i : s) cout << i << ' ';
    cout << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int T;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  