#include <bits/stdc++.h>
using namespace std;
long long int mul(long long int a, long long int b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long int add(long long int a, long long int b) {
  return (a % 1000000007 + b % 1000000007) % 1000000007;
}
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 extgcd(long long int a, long long int b, long long int& x,
                     long long int& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long int x1, y1;
  long long int d = extgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool isPowerOfTwo(long long int n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
vector<long long int> factors(long long int n) {
  vector<long long int> f;
  for (long long int x = 2; x * x <= n; x++) {
    while (n % x == 0) {
      f.push_back(x);
      n /= x;
    }
  }
  if (n > 1) f.push_back(n);
  return f;
}
long long int bin_exp_mod(long long int a, long long int n) {
  long long int res = 1;
  a %= 1000000007;
  if (a == 0) return 0;
  while (n > 0) {
    if (n & 1) res = ((res % 1000000007) * (a % 1000000007)) % 1000000007;
    n = n >> 1;
    a = ((a % 1000000007) * (a % 1000000007)) % 1000000007;
  }
  res %= 1000000007;
  return res;
}
void solve() {
  long long int a, b;
  cin >> a >> b;
  long long int n = a + b;
  set<long long int> st;
  if ((a + b) % 2 == 0) {
    long long int x = (a + b) / 2;
    long long int y = (a + b) / 2;
    long long int d = abs(a - b) / 2;
    for (long long int i = d; i < a + b - d + 1; i++) {
      st.insert(i);
      i++;
    }
  } else {
    long long int x = (a + b) / 2 + 1;
    long long int y = (a + b) / 2;
    long long int d = abs(a - b) / 2;
    for (long long int i = d; i < a + b - d + 1; i++) {
      st.insert(i);
    }
  }
  cout << st.size() << "\n";
  for (auto x : st) cout << x << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### 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>
using namespace std;
long long int mul(long long int a, long long int b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long int add(long long int a, long long int b) {
  return (a % 1000000007 + b % 1000000007) % 1000000007;
}
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 extgcd(long long int a, long long int b, long long int& x,
                     long long int& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long int x1, y1;
  long long int d = extgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool isPowerOfTwo(long long int n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
vector<long long int> factors(long long int n) {
  vector<long long int> f;
  for (long long int x = 2; x * x <= n; x++) {
    while (n % x == 0) {
      f.push_back(x);
      n /= x;
    }
  }
  if (n > 1) f.push_back(n);
  return f;
}
long long int bin_exp_mod(long long int a, long long int n) {
  long long int res = 1;
  a %= 1000000007;
  if (a == 0) return 0;
  while (n > 0) {
    if (n & 1) res = ((res % 1000000007) * (a % 1000000007)) % 1000000007;
    n = n >> 1;
    a = ((a % 1000000007) * (a % 1000000007)) % 1000000007;
  }
  res %= 1000000007;
  return res;
}
void solve() {
  long long int a, b;
  cin >> a >> b;
  long long int n = a + b;
  set<long long int> st;
  if ((a + b) % 2 == 0) {
    long long int x = (a + b) / 2;
    long long int y = (a + b) / 2;
    long long int d = abs(a - b) / 2;
    for (long long int i = d; i < a + b - d + 1; i++) {
      st.insert(i);
      i++;
    }
  } else {
    long long int x = (a + b) / 2 + 1;
    long long int y = (a + b) / 2;
    long long int d = abs(a - b) / 2;
    for (long long int i = d; i < a + b - d + 1; i++) {
      st.insert(i);
    }
  }
  cout << st.size() << "\n";
  for (auto x : st) cout << x << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```