#include <bits/stdc++.h>
using namespace std;
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << '\n';
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
void run() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
}
const int N = 1e5 + 5, mod = 1e9 + 7, mod2 = 998244353, MAXN = 100000 + 29;
void files() {}
int arr[N], n;
bool go(bool f) {
  if (f)
    for (int i = 0; i < n - 1; i += 2) {
      if (arr[i] > arr[i + 1]) swap(arr[i], arr[i + 1]);
    }
  else
    for (int i = 1; i < n - 1; i += 2) {
      if (arr[i] > arr[i + 1]) swap(arr[i], arr[i + 1]);
    }
  bool ret = true;
  for (int i = 0; i < n; ++i) {
    ret &= arr[i] == (i + 1);
  }
  return ret;
}
int a, b;
bool solve(int i, int j, int k, bool t) {
  if (i < 0 || j < 0) return 0;
  if (i == 0 && j == 0) {
    return k == 0;
  }
  if (k < 0) return 0;
  if (t) {
    return solve(i - 1, j, k, t ^ 1) || solve(i, j - 1, k - 1, t ^ 1);
  } else
    return solve(i, j - 1, k, t ^ 1) || solve(i - 1, j, k - 1, t ^ 1);
}
void solve() {
  cin >> a >> b;
  int diff = abs(a - b);
  vector<int> v;
  for (int i = diff / 2; i <= a + b - diff / 2;) {
    v.push_back(i);
    i += 1 + (diff % 2 == 0);
  }
  cout << v.size() << '\n';
  for (int i = 0; i < v.size(); ++i) {
    cout << v[i] << ' ';
  }
  cout << '\n';
}
int main() {
  files();
  int t = 1;
  cin >> t;
  while (t--) solve();
}

### Prompt

Create a solution in Cpp for 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;
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << '\n';
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
void run() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
}
const int N = 1e5 + 5, mod = 1e9 + 7, mod2 = 998244353, MAXN = 100000 + 29;
void files() {}
int arr[N], n;
bool go(bool f) {
  if (f)
    for (int i = 0; i < n - 1; i += 2) {
      if (arr[i] > arr[i + 1]) swap(arr[i], arr[i + 1]);
    }
  else
    for (int i = 1; i < n - 1; i += 2) {
      if (arr[i] > arr[i + 1]) swap(arr[i], arr[i + 1]);
    }
  bool ret = true;
  for (int i = 0; i < n; ++i) {
    ret &= arr[i] == (i + 1);
  }
  return ret;
}
int a, b;
bool solve(int i, int j, int k, bool t) {
  if (i < 0 || j < 0) return 0;
  if (i == 0 && j == 0) {
    return k == 0;
  }
  if (k < 0) return 0;
  if (t) {
    return solve(i - 1, j, k, t ^ 1) || solve(i, j - 1, k - 1, t ^ 1);
  } else
    return solve(i, j - 1, k, t ^ 1) || solve(i - 1, j, k - 1, t ^ 1);
}
void solve() {
  cin >> a >> b;
  int diff = abs(a - b);
  vector<int> v;
  for (int i = diff / 2; i <= a + b - diff / 2;) {
    v.push_back(i);
    i += 1 + (diff % 2 == 0);
  }
  cout << v.size() << '\n';
  for (int i = 0; i < v.size(); ++i) {
    cout << v[i] << ' ';
  }
  cout << '\n';
}
int main() {
  files();
  int t = 1;
  cin >> t;
  while (t--) solve();
}

```