#include <bits/stdc++.h>
using namespace std;
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &i : v) in >> i;
  return in;
}
template <typename A, typename B>
istream &operator>>(istream &in, pair<A, B> &a) {
  in >> a.first >> a.second;
  return in;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long l, long long r) {
  uniform_int_distribution<long long> uid(l, r);
  return uid(rng);
}
const string no = "No";
const string yes = "Yes";
void solve() {
  long long a, b;
  cin >> a >> b;
  long long rounds = a + b;
  long long matcha = (rounds + 1) / 2, matchb = rounds - matcha;
  set<long long> s;
  for (long long i = 0; i <= matchb; i++) {
    long long pointsa = matcha + i;
    long long pointsb = matchb - i;
    long long breaksfroma = pointsa - a;
    pointsa -= breaksfroma;
    pointsb += breaksfroma;
    if (pointsa < i) {
      long long takeback = i - pointsa;
      breaksfroma -= takeback;
      pointsa += takeback;
      pointsb += takeback;
    };
    ;
    if (b == pointsb && a == pointsa && breaksfroma >= 0) {
      s.insert(i + breaksfroma);
    }
  };
  matcha = rounds / 2, matchb = rounds - matcha;
  for (long long i = 0; i <= matcha; i++) {
    long long pointsb = matchb + i;
    long long pointsa = matcha - i;
    long long breaksfromb = pointsb - b;
    pointsb -= breaksfromb;
    pointsa += breaksfromb;
    if (pointsb < i) {
      long long takeback = i - pointsb;
      breaksfromb -= takeback;
      pointsb += takeback;
      pointsa -= takeback;
    };
    ;
    if (b == pointsb && a == pointsa && breaksfromb >= 0) {
      s.insert(i + breaksfromb);
    }
  };
  cout << (long long)(s).size() << '\n';
  for (long long i : s) {
    cout << i << ' ';
  }
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  for (long long tc = 1; tc <= t; tc++) {
    ;
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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;
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &i : v) in >> i;
  return in;
}
template <typename A, typename B>
istream &operator>>(istream &in, pair<A, B> &a) {
  in >> a.first >> a.second;
  return in;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long l, long long r) {
  uniform_int_distribution<long long> uid(l, r);
  return uid(rng);
}
const string no = "No";
const string yes = "Yes";
void solve() {
  long long a, b;
  cin >> a >> b;
  long long rounds = a + b;
  long long matcha = (rounds + 1) / 2, matchb = rounds - matcha;
  set<long long> s;
  for (long long i = 0; i <= matchb; i++) {
    long long pointsa = matcha + i;
    long long pointsb = matchb - i;
    long long breaksfroma = pointsa - a;
    pointsa -= breaksfroma;
    pointsb += breaksfroma;
    if (pointsa < i) {
      long long takeback = i - pointsa;
      breaksfroma -= takeback;
      pointsa += takeback;
      pointsb += takeback;
    };
    ;
    if (b == pointsb && a == pointsa && breaksfroma >= 0) {
      s.insert(i + breaksfroma);
    }
  };
  matcha = rounds / 2, matchb = rounds - matcha;
  for (long long i = 0; i <= matcha; i++) {
    long long pointsb = matchb + i;
    long long pointsa = matcha - i;
    long long breaksfromb = pointsb - b;
    pointsb -= breaksfromb;
    pointsa += breaksfromb;
    if (pointsb < i) {
      long long takeback = i - pointsb;
      breaksfromb -= takeback;
      pointsb += takeback;
      pointsa -= takeback;
    };
    ;
    if (b == pointsb && a == pointsa && breaksfromb >= 0) {
      s.insert(i + breaksfromb);
    }
  };
  cout << (long long)(s).size() << '\n';
  for (long long i : s) {
    cout << i << ' ';
  }
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  for (long long tc = 1; tc <= t; tc++) {
    ;
    solve();
  }
  return 0;
}

```