#include <bits/stdc++.h>
using namespace std;
void main2() {
  long long int a, b, n, t, x, p, q;
  cin >> a >> b;
  if (a < b) swap(a, b);
  vector<long long int> ans;
  t = (a + b) / 2;
  if ((a + b) % 2 == 1) t++;
  p = min(t, a);
  q = min(a + b - t, b);
  for (long long int i = 0; i <= q; i++) ans.push_back(p - q + 2 * i);
  if ((a + b) % 2 == 1) {
    p = min(a + b - t, a);
    q = min(t, b);
    if (q > p) swap(p, q);
    for (long long int i = 0; i <= q; i++) ans.push_back(p - q + 2 * i);
  }
  cout << ans.size() << "\n";
  sort(ans.begin(), ans.end());
  for (long long int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    main2();
  }
  return 0;
}

### Prompt

Generate a Cpp solution 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;
void main2() {
  long long int a, b, n, t, x, p, q;
  cin >> a >> b;
  if (a < b) swap(a, b);
  vector<long long int> ans;
  t = (a + b) / 2;
  if ((a + b) % 2 == 1) t++;
  p = min(t, a);
  q = min(a + b - t, b);
  for (long long int i = 0; i <= q; i++) ans.push_back(p - q + 2 * i);
  if ((a + b) % 2 == 1) {
    p = min(a + b - t, a);
    q = min(t, b);
    if (q > p) swap(p, q);
    for (long long int i = 0; i <= q; i++) ans.push_back(p - q + 2 * i);
  }
  cout << ans.size() << "\n";
  sort(ans.begin(), ans.end());
  for (long long int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    main2();
  }
  return 0;
}

```