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. 
#include <bits/stdc++.h>
using namespace std;
long long is_valid(long long a, long long b, long long ex, long long k) {
  long long n = a + b;
  long long c_a = n / 2;
  long long c_b = n / 2;
  if (n % 2) {
    if (ex)
      c_a++;
    else
      c_b++;
  }
  long long x = a + k - c_a;
  if (x % 2) return 0;
  if (x < 0) return 0;
  x /= 2;
  if (k - x <= c_a and x <= c_b and c_b + k - 2 * x == b and
      2 * x - k + c_a == a) {
    return 1;
  }
  return 0;
}
void solve(long long tc) {
  long long a, b;
  cin >> a >> b;
  vector<long long> sol;
  for (long long i = 0; i <= a + b; i++) {
    if (is_valid(a, b, 1, i) or is_valid(a, b, 0, i)) {
      sol.push_back(i);
    }
  }
  cout << sol.size() << endl;
  for (long long i = 0; i < sol.size(); i++) {
    cout << sol[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long t = 1;
  long long tc = 1;
  cin >> t;
  while (t--) {
    solve(tc);
    tc++;
  }
  return 0;
}
