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>
  #pragma GCC optimize("O3")
  #pragma GCC target("sse4")
  using namespace std;
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <typename T, size_t N>
  long long SIZE(const T (&t)[N]) {
    return N;
  }
  template <typename T>
  long long SIZE(const T& t) {
    return t.size();
  }
  string to_string(const string s, long long x1 = 0, long long x2 = 1e9) {
    return '"' + ((x1 < s.size()) ? s.substr(x1, x2 - x1 + 1) : "") + '"';
  }
  string to_string(const char* s) { return to_string((string)s); }
  string to_string(const bool b) { return (b ? "true" : "false"); }
  string to_string(const char c) { return string({c}); }
  template <size_t N>
  string to_string(const bitset<N>& b, long long x1 = 0, long long x2 = 1e9) {
    string t = "";
    for (long long __iii__ = min(x1, SIZE(b)), __jjj__ = min(x2, SIZE(b) - 1);
         __iii__ <= __jjj__; ++__iii__) {
      t += b[__iii__] + '0';
    }
    return '"' + t + '"';
  }
  template <typename A, typename... C>
  string to_string(const A(&v), long long x1 = 0, long long x2 = 1e9,
                   C... coords);
  long long l_v_l_v_l = 0, t_a_b_s = 0;
  template <typename A, typename B>
  string to_string(const pair<A, B>& p) {
    l_v_l_v_l++;
    string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
    l_v_l_v_l--;
    return res;
  }
  template <typename A, typename... C>
  string to_string(const A(&v), long long x1, long long x2, C... coords) {
    long long rnk = rank<A>::value;
    string tab(t_a_b_s, ' ');
    string res = "";
    bool first = true;
    if (l_v_l_v_l == 0) res += '\n';
    res += tab + "[";
    x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v));
    auto l = begin(v);
    advance(l, x1);
    auto r = l;
    advance(r, (x2 - x1) + (x2 < SIZE(v)));
    for (auto e = l; e != r; e = next(e)) {
      if (!first) {
        res += ", ";
      }
      first = false;
      l_v_l_v_l++;
      if (e != l) {
        if (rnk > 1) {
          res += '\n';
          t_a_b_s = l_v_l_v_l;
        };
      } else {
        t_a_b_s = 0;
      }
      res += to_string(*e, coords...);
      l_v_l_v_l--;
    }
    res += "]";
    if (l_v_l_v_l == 0) res += '\n';
    return res;
  }
  void dbgm() { ; }
  template <typename Heads, typename... Tails>
  void dbgm(Heads H, Tails... T) {
    cout << to_string(H) << " | ";
    dbgm(T...);
  }
  long long msb(long long x);
  long long lsb(long long x);
  vector<long long> tovec(string a);
  string tostr(vector<long long> a);
  const long long md = 1e9 + 7, MX = 500001;
  long long get(long long n, long long a, long long b, long long colored) {
    if (colored > a) return -1;
    long long even = (n + 1) / 2;
    long long odd = n / 2;
    if (colored > even) return -1;
    if (even - colored > b) return -1;
    return even - colored + a - colored;
  }
  void solve() {
    long long a, b;
    cin >> a >> b;
    set<long long> s;
    for (long long i = 0; i <= a; i++) {
      long long x = get(a + b, a, b, i);
      if (x != -1) s.insert(x);
    }
    for (long long i = 0; i <= b; i++) {
      long long x = get(a + b, b, a, i);
      if (x != -1) s.insert(x);
    }
    cout << s.size() << '\n';
    for (long long x : s) cout << x << " ";
    return;
  }
  signed main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    long long T = 1;
    cin >> T;
    while (T--) {
      solve();
      if (T) cout << '\n';
    }
    return 0;
  }
  long long msb(long long x) {
    for (long long i = 60; i > -1; i--)
      if ((1LL << i) & x) return i;
    return -1;
  }
  long long lsb(long long x) {
    for (long long i = 0; i < 61; i++)
      if ((1LL << i) & x) return i;
    return -1;
  }
  vector<long long> tovec(string a) {
    vector<long long> ans;
    for (char x : a) ans.push_back(((long long)(x) - 'a'));
    return ans;
  }
  string tostr(vector<long long> a) {
    string ans;
    for (long long x : a) ans += ((char)((x) + 'a'));
    return ans;
  }
  