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;
void solve() {
  long long alice, bob;
  cin >> alice >> bob;
  long long n = alice + bob;
  if (n == 1) {
    cout << "2\n0 1\n";
    return;
  }
  set<long long> ans;
  {
    long long a = alice;
    long long b = bob;
    long long p = min(a, b);
    long long a_end = 1e9;
    long long b_end = 1e9;
    for (long long i = 1; i <= n; i++) {
      if (i % 2) {
        if (b) {
          b--;
        } else {
          a--;
        }
        if (!b) {
          b_end = min(b_end, i);
        }
      } else {
        if (a) {
          a--;
        } else {
          b--;
        }
        if (!a) {
          a_end = min(a_end, i);
        }
      }
    }
    long long st = (n - 2 * p) / 2 + (a_end < b_end);
    for (long long k = st; k <= st + 2 * p; k += 2) {
      ans.insert(k);
    }
  }
  {
    long long a = alice;
    long long b = bob;
    long long p = min(a, b);
    long long a_end = 1e9;
    long long b_end = 1e9;
    for (long long i = 1; i <= n; i++) {
      if (!(i % 2)) {
        if (b) {
          b--;
        } else {
          a--;
        }
        if (!b) {
          b_end = min(b_end, i);
        }
      } else {
        if (a) {
          a--;
        } else {
          b--;
        }
        if (!a) {
          a_end = min(a_end, i);
        }
      }
    }
    long long st = (n - 2 * p) / 2 + (a_end > b_end);
    for (long long k = st; k <= st + 2 * p; k += 2) {
      ans.insert(k);
    }
  }
  cout << ans.size() << '\n';
  for (auto& x : ans) {
    cout << x << ' ';
  }
  cout << '\n';
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
