#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
long long qpow(long long b, long long p, long long Mod = 1e9 + 7) {
  long long r = 1;
  while (p) {
    if (p & 1) r = r * b % Mod;
    b = b * b % Mod;
    p >>= 1;
  }
  return r;
}
const int mod = 1e9 + 7;
const int N = 2e7 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = (1ll << 60);
void pre(int n = 1e7) { return; }
void solves() {
  int a, b;
  cin >> a >> b;
  vector<int> ans;
  if ((a + b) & 1) {
    for (int i = abs(a - b) / 2; i <= (a + b + 1) / 2 + min(a, b); ++i) {
      ans.push_back(i);
    }
  } else {
    for (int i = abs(a - b) / 2; i <= (a + b) / 2 + min(a, b); i += 2)
      ans.push_back(i);
  }
  cout << ans.size() << '\n';
  for (auto& it : ans) cout << it << " ";
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int _OwO_ = 1;
  cin >> _OwO_;
  pre();
  while (_OwO_--) solves();
}

### Prompt

Please create a solution in cpp 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>
#pragma GCC optimize(2)
using namespace std;
long long qpow(long long b, long long p, long long Mod = 1e9 + 7) {
  long long r = 1;
  while (p) {
    if (p & 1) r = r * b % Mod;
    b = b * b % Mod;
    p >>= 1;
  }
  return r;
}
const int mod = 1e9 + 7;
const int N = 2e7 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = (1ll << 60);
void pre(int n = 1e7) { return; }
void solves() {
  int a, b;
  cin >> a >> b;
  vector<int> ans;
  if ((a + b) & 1) {
    for (int i = abs(a - b) / 2; i <= (a + b + 1) / 2 + min(a, b); ++i) {
      ans.push_back(i);
    }
  } else {
    for (int i = abs(a - b) / 2; i <= (a + b) / 2 + min(a, b); i += 2)
      ans.push_back(i);
  }
  cout << ans.size() << '\n';
  for (auto& it : ans) cout << it << " ";
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int _OwO_ = 1;
  cin >> _OwO_;
  pre();
  while (_OwO_--) solves();
}

```