#include <bits/stdc++.h>
using namespace std;
const int OO = 1e9;
const double EPS = 1e-9;
template <class T>
void _db(const char* dbStr, T e) {
  cout << dbStr << " = " << e << endl;
}
template <class T, class... L>
void _db(const char* dbStr, T e, L... r) {
  while (*dbStr != ',') cout << *dbStr++;
  cout << " = " << e << ',';
  _db(dbStr + 1, r...);
}
template <class S, class T>
ostream& operator<<(ostream& o, const map<S, T>& v) {
  o << "[";
  int i = 0;
  for (const pair<S, T>& pr : v)
    o << (!i++ ? "" : ", ") << "{" << pr.first << " : " << pr.second << "}";
  return o << "]";
}
template <template <class, class...> class S, class T, class... L>
ostream& operator<<(ostream& o, const S<T, L...>& v) {
  o << "[";
  int i = 0;
  for (const auto& e : v) o << (!i++ ? "" : ", ") << e;
  return o << "]";
}
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& pr) {
  return o << "(" << pr.first << ", " << pr.second << ")";
}
ostream& operator<<(ostream& o, const string& s) {
  for (const char& c : s) o << c;
  return o;
}
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;
template <class T>
using VVV = VV<V<T>>;
using ll = long long;
using pii = pair<int, int>;
using vi = V<int>;
using vii = V<pii>;
using vvi = VV<int>;
using mii = map<int, int>;
using umii = unordered_map<int, int>;
using si = set<int>;
using usi = unordered_set<int>;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  cin.tie(0);
  int T;
  cin >> T;
  for (int tc = 1; tc <= T; ++tc) {
    int a, b;
    cin >> a >> b;
    si ans;
    for (int I = 0; I <= 1; ++I) {
      int odd = (a + b + 1) / 2, even = (a + b) / 2;
      for (int x = 0; x <= min(odd, a); ++x) {
        int rem = a - x;
        if (rem > even) continue;
        ans.insert((odd - x) + rem);
      }
      swap(a, b);
    }
    cout << int(ans.size()) << '\n';
    for (auto& elem : ans) {
      cout << elem << ' ';
    }
    cout << '\n';
  }
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int OO = 1e9;
const double EPS = 1e-9;
template <class T>
void _db(const char* dbStr, T e) {
  cout << dbStr << " = " << e << endl;
}
template <class T, class... L>
void _db(const char* dbStr, T e, L... r) {
  while (*dbStr != ',') cout << *dbStr++;
  cout << " = " << e << ',';
  _db(dbStr + 1, r...);
}
template <class S, class T>
ostream& operator<<(ostream& o, const map<S, T>& v) {
  o << "[";
  int i = 0;
  for (const pair<S, T>& pr : v)
    o << (!i++ ? "" : ", ") << "{" << pr.first << " : " << pr.second << "}";
  return o << "]";
}
template <template <class, class...> class S, class T, class... L>
ostream& operator<<(ostream& o, const S<T, L...>& v) {
  o << "[";
  int i = 0;
  for (const auto& e : v) o << (!i++ ? "" : ", ") << e;
  return o << "]";
}
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& pr) {
  return o << "(" << pr.first << ", " << pr.second << ")";
}
ostream& operator<<(ostream& o, const string& s) {
  for (const char& c : s) o << c;
  return o;
}
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;
template <class T>
using VVV = VV<V<T>>;
using ll = long long;
using pii = pair<int, int>;
using vi = V<int>;
using vii = V<pii>;
using vvi = VV<int>;
using mii = map<int, int>;
using umii = unordered_map<int, int>;
using si = set<int>;
using usi = unordered_set<int>;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  cin.tie(0);
  int T;
  cin >> T;
  for (int tc = 1; tc <= T; ++tc) {
    int a, b;
    cin >> a >> b;
    si ans;
    for (int I = 0; I <= 1; ++I) {
      int odd = (a + b + 1) / 2, even = (a + b) / 2;
      for (int x = 0; x <= min(odd, a); ++x) {
        int rem = a - x;
        if (rem > even) continue;
        ans.insert((odd - x) + rem);
      }
      swap(a, b);
    }
    cout << int(ans.size()) << '\n';
    for (auto& elem : ans) {
      cout << elem << ' ';
    }
    cout << '\n';
  }
  return 0;
}

```