#include <bits/stdc++.h>
using namespace std;
constexpr long long MOD = 998244353;
constexpr long long mod = 1e9 + 7;
const double eps = 1e-7;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
const long long N = 3e6 + 5;
vector<long long> adj[N];
long long power(long long n, long long p) {
  if (p == 0) return 1;
  if (p == 1) return n;
  if (p % 2)
    return power(n, p - 1) * n;
  else {
    long long x = power(n, p / 2);
    return x * x;
  }
}
long long modpow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
long long nsum(long long num) { return (num * (num + 1)) / 2; }
void edge(long long u, long long v) {
  adj[u].push_back(v);
  adj[v].push_back(u);
}
void solve() {
  long long n, a, b;
  string s;
  cin >> a >> b;
  long long all = a + b;
  set<long long> st;
  long long aw = 0, ah = 0, bw = 0, bh = 0;
  for (long long i = 0; i <= all / 2; i++) {
    if (i > a) break;
    aw = i;
    ah = a - aw;
    bh = all / 2 - aw;
    bw = b - bh;
    if (ah < 0 || bh < 0 || aw < 0 || bw < 0) continue;
    if ((ah + aw) == a && (bh + bw) == b) {
      st.insert(aw + bw);
    }
  }
  for (long long i = 0; i <= (all + 1) / 2; i++) {
    if (i > a) break;
    aw = i;
    ah = a - aw;
    bh = (all + 1) / 2 - aw;
    bw = b - bh;
    if (ah < 0 || bh < 0 || aw < 0 || bw < 0) continue;
    if ((ah + aw) == a && (bh + bw) == b) {
      st.insert(aw + bw);
    }
  }
  cout << st.size() << "\n";
  for (auto it : st) cout << it << ' ';
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long Test = 1;
  cin >> Test;
  while (Test--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
constexpr long long MOD = 998244353;
constexpr long long mod = 1e9 + 7;
const double eps = 1e-7;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
const long long N = 3e6 + 5;
vector<long long> adj[N];
long long power(long long n, long long p) {
  if (p == 0) return 1;
  if (p == 1) return n;
  if (p % 2)
    return power(n, p - 1) * n;
  else {
    long long x = power(n, p / 2);
    return x * x;
  }
}
long long modpow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
long long nsum(long long num) { return (num * (num + 1)) / 2; }
void edge(long long u, long long v) {
  adj[u].push_back(v);
  adj[v].push_back(u);
}
void solve() {
  long long n, a, b;
  string s;
  cin >> a >> b;
  long long all = a + b;
  set<long long> st;
  long long aw = 0, ah = 0, bw = 0, bh = 0;
  for (long long i = 0; i <= all / 2; i++) {
    if (i > a) break;
    aw = i;
    ah = a - aw;
    bh = all / 2 - aw;
    bw = b - bh;
    if (ah < 0 || bh < 0 || aw < 0 || bw < 0) continue;
    if ((ah + aw) == a && (bh + bw) == b) {
      st.insert(aw + bw);
    }
  }
  for (long long i = 0; i <= (all + 1) / 2; i++) {
    if (i > a) break;
    aw = i;
    ah = a - aw;
    bh = (all + 1) / 2 - aw;
    bw = b - bh;
    if (ah < 0 || bh < 0 || aw < 0 || bw < 0) continue;
    if ((ah + aw) == a && (bh + bw) == b) {
      st.insert(aw + bw);
    }
  }
  cout << st.size() << "\n";
  for (auto it : st) cout << it << ' ';
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long Test = 1;
  cin >> Test;
  while (Test--) {
    solve();
  }
  return 0;
}

```