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;
  long long int inf = 1e18;
  long long int mod = 1e9 + 7;
  struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
      x += 0x9e3779b97f4a7c15;
      x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
      x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
      return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
      static const uint64_t FIXED_RANDOM =
          chrono::steady_clock::now().time_since_epoch().count();
      return splitmix64(x + FIXED_RANDOM);
    }
  };
  long long int power(long long int x, long long int y, long long int p) {
    if (x == 0) return 0;
    long long int res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  long long int pwr(long long int x, long long int y) {
    if (x == 0) return 0;
    long long int res = 1;
    x = x;
    while (y > 0) {
      if (y & 1) res = (res * x);
      y = y >> 1;
      x = (x * x);
    }
    return res;
  }
  long long int modInverse(long long int n, long long int p) {
    return power(n, p - 2, p);
  }
  long long int ncr(long long int n, long long int r) {
    if (r > n) {
      return 0;
    }
    if (r > n - r) {
      r = n - r;
    }
    long long int ans = 1;
    for (long long int i = 1; i <= r; i++) {
      ans *= (n - i + 1);
      ans %= mod;
      ans *= modInverse(i, mod);
      ans %= mod;
    }
    return ans;
  }
  long long int gcd(long long int a, long long int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int TESTS, i, j, k, flag, t1, t2, t3, t4, t5, t6, t7, t8, t9;
    cin >> TESTS;
    while (TESTS--) {
      long long int a, b;
      cin >> a >> b;
      t1 = min(a, b);
      long long int left = (a + b) - (2 * t1);
      set<long long int> out;
      t2 = left / 2;
      out.insert(t2);
      out.insert(a + b - t2);
      for (i = 1; i <= t1; i++) {
        t2 += 2;
        out.insert(t2);
        out.insert(a + b - t2);
      }
      cout << out.size() << "\n";
      for (auto c : out) cout << c << " ";
      cout << "\n";
    }
  }
  