#include <bits/stdc++.h>
using namespace std;
const long double PI = 3.14159265359;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long randint(long long a, long long b) {
  return uniform_int_distribution<long long>(a, b)(rng);
}
template <typename... Args>
void logger(string vars, Args&&... values) {
  cout << vars << " = ";
  string delim = "";
  (..., (cout << delim << values, delim = ", "));
  cout << endl;
}
void print(vector<int>& v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    cout << v[i];
    if (i + 1 != v.size()) cout << ", ";
  }
  cout << "]" << endl;
}
void print(pair<int, int>& p) {
  cout << "{" << p.first << ", " << p.second << "}" << endl;
}
vector<int> junta(pair<int, int> a, pair<int, int> b) {
  int mini = 0;
  int maxi = 1;
  if (a.first > b.first) {
    mini = a.first - b.first;
    maxi += a.second;
  } else {
    mini = a.second - b.second;
    maxi += a.first;
  }
  vector<int> v;
  int cont = 0;
  for (int i = mini; i <= a.first + a.second && (cont++) < maxi; i += 2) {
    v.push_back(i);
  }
  return v;
}
signed main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int a, b;
    cin >> a >> b;
    int n = (a + b);
    pair<int, int> t1 = {n / 2, (n + 1) / 2};
    pair<int, int> t2 = {(n + 1) / 2, n / 2};
    auto v1 = junta({a, b}, t1);
    auto v2 = junta({a, b}, t2);
    for (auto k : v2) {
      v1.push_back(k);
    }
    sort(v1.begin(), v1.end());
    auto un = unique(v1.begin(), v1.end());
    int cont = 0;
    for (auto i = v1.begin(); i != un; i++) {
      cont++;
    }
    cout << cont << '\n';
    for (auto i = v1.begin(); i != un; i++) {
      cout << *i << ' ';
    }
    cout << '\n';
  }
}

### Prompt

Please create a solution in Cpp to 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 long double PI = 3.14159265359;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long randint(long long a, long long b) {
  return uniform_int_distribution<long long>(a, b)(rng);
}
template <typename... Args>
void logger(string vars, Args&&... values) {
  cout << vars << " = ";
  string delim = "";
  (..., (cout << delim << values, delim = ", "));
  cout << endl;
}
void print(vector<int>& v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    cout << v[i];
    if (i + 1 != v.size()) cout << ", ";
  }
  cout << "]" << endl;
}
void print(pair<int, int>& p) {
  cout << "{" << p.first << ", " << p.second << "}" << endl;
}
vector<int> junta(pair<int, int> a, pair<int, int> b) {
  int mini = 0;
  int maxi = 1;
  if (a.first > b.first) {
    mini = a.first - b.first;
    maxi += a.second;
  } else {
    mini = a.second - b.second;
    maxi += a.first;
  }
  vector<int> v;
  int cont = 0;
  for (int i = mini; i <= a.first + a.second && (cont++) < maxi; i += 2) {
    v.push_back(i);
  }
  return v;
}
signed main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int a, b;
    cin >> a >> b;
    int n = (a + b);
    pair<int, int> t1 = {n / 2, (n + 1) / 2};
    pair<int, int> t2 = {(n + 1) / 2, n / 2};
    auto v1 = junta({a, b}, t1);
    auto v2 = junta({a, b}, t2);
    for (auto k : v2) {
      v1.push_back(k);
    }
    sort(v1.begin(), v1.end());
    auto un = unique(v1.begin(), v1.end());
    int cont = 0;
    for (auto i = v1.begin(); i != un; i++) {
      cont++;
    }
    cout << cont << '\n';
    for (auto i = v1.begin(); i != un; i++) {
      cout << *i << ' ';
    }
    cout << '\n';
  }
}

```