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 maxn = 5e5 + 5;
  int s[maxn];
  void solve() {
    long long n, m, i, j, f = 1, k, b, a, t = 0, c;
    scanf("%lld%lld", &a, &b);
    n = a + b;
    a = min(a, b);
    c = n / 2;
    if (n % 2) {
      for (i = n - (c + 1 + a);; i += 2) {
        s[++t] = i;
        if (t == a + 1) break;
      }
      for (i = 1; i <= t; i++)
        if (s[i] > c) s[i] = n - s[i];
      printf("%d\n", t * 2);
      sort(s + 1, s + t + 1);
      for (i = 1; i <= t; i++) printf("%d ", s[i]);
      for (i = t; i >= 1; i--) printf("%d ", n - s[i]);
      printf("\n");
    } else {
      for (i = n - (c + a); i <= c; i += 2) {
        s[++t] = i;
      }
      if (s[t] == c) {
        printf("%d\n", t * 2 - 1);
        for (i = 1; i < t; i++) printf("%d ", s[i]);
        for (i = t; i >= 1; i--) printf("%d ", n - s[i]);
      } else {
        printf("%d\n", t * 2);
        for (i = 1; i <= t; i++) printf("%d ", s[i]);
        for (i = t; i >= 1; i--) printf("%d ", n - s[i]);
      }
      printf("\n");
    }
  }
  int main() {
    int T = 1;
    scanf("%d", &T);
    while (T--) solve();
    return 0;
  }
  