#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int result = power(a, b / 2);
  if (b & 1) return result * result * a;
  return result * result;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int a, b;
    cin >> a >> b;
    if (a == 0 or b == 0) {
      if (max(a, b) % 2) {
        cout << 2 << "\n";
        cout << max(a, b) / 2 << " " << max(a, b) / 2 + 1 << "\n";
      } else {
        cout << 1 << "\n";
        cout << max(a, b) / 2 << "\n";
      }
    } else {
      long long int gap = abs(a - b);
      if (gap == 1) {
        cout << a + b + 1 << "\n";
        for (long long int i = 0; i <= a + b; i++) {
          cout << i << " ";
        }
        cout << "\n";
      } else if (gap % 2 == 0) {
        vector<long long int> v;
        long long int ans = 0;
        long long int total = (a + b) / 2;
        long long int start = max(a, b) - (a + b) / 2;
        long long int end =
            (a + b) / 2 + ((a + b) / 2 - (max(a, b) - (a + b) / 2));
        for (long long int i = start; i <= end; i += 2) {
          v.push_back(i);
          ans++;
        }
        cout << ans << "\n";
        for (long long int i = 0; i < v.size(); i++) {
          cout << v[i] << " ";
        }
        cout << "\n";
      } else {
        long long int ans1 = 0;
        long long int start = max(a, b) - (a + b) / 2 - 1;
        long long int end = (a + b) / 2 + 1 + (a + b) / 2 - start;
        vector<long long int> v1;
        for (long long int i = start; i <= end; i++) {
          v1.push_back(i);
        }
        cout << v1.size() << "\n";
        for (long long int i = 0; i < v1.size(); i++) {
          cout << v1[i] << " ";
        }
        cout << "\n";
      }
    }
  }
}

### 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;
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int result = power(a, b / 2);
  if (b & 1) return result * result * a;
  return result * result;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int a, b;
    cin >> a >> b;
    if (a == 0 or b == 0) {
      if (max(a, b) % 2) {
        cout << 2 << "\n";
        cout << max(a, b) / 2 << " " << max(a, b) / 2 + 1 << "\n";
      } else {
        cout << 1 << "\n";
        cout << max(a, b) / 2 << "\n";
      }
    } else {
      long long int gap = abs(a - b);
      if (gap == 1) {
        cout << a + b + 1 << "\n";
        for (long long int i = 0; i <= a + b; i++) {
          cout << i << " ";
        }
        cout << "\n";
      } else if (gap % 2 == 0) {
        vector<long long int> v;
        long long int ans = 0;
        long long int total = (a + b) / 2;
        long long int start = max(a, b) - (a + b) / 2;
        long long int end =
            (a + b) / 2 + ((a + b) / 2 - (max(a, b) - (a + b) / 2));
        for (long long int i = start; i <= end; i += 2) {
          v.push_back(i);
          ans++;
        }
        cout << ans << "\n";
        for (long long int i = 0; i < v.size(); i++) {
          cout << v[i] << " ";
        }
        cout << "\n";
      } else {
        long long int ans1 = 0;
        long long int start = max(a, b) - (a + b) / 2 - 1;
        long long int end = (a + b) / 2 + 1 + (a + b) / 2 - start;
        vector<long long int> v1;
        for (long long int i = start; i <= end; i++) {
          v1.push_back(i);
        }
        cout << v1.size() << "\n";
        for (long long int i = 0; i < v1.size(); i++) {
          cout << v1[i] << " ";
        }
        cout << "\n";
      }
    }
  }
}

```