#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const double EXP = exp(1.0);
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long lowbit(long long x) { return x & (-x); }
stack<int> st;
int arr[maxn];
bool check(int brk, int a, int b) {
  int tot = a + b;
  if (tot & 1) {
    int tmpa = (tot + 1) / 2;
    int tmpb = (tot) / 2;
    for (int i = 0; i <= brk; i++) {
      tmpa = (tot + 1) / 2;
      tmpb = (tot) / 2;
      int brka = i, brkb = brk - i;
      if (brka > tmpb || brkb > tmpa) continue;
      tmpa = tmpa + i - brkb;
      tmpb = tmpb + brkb - i;
      if (tmpa < 0 || tmpb < 0) continue;
      if (a == tmpa && b == tmpb) {
        return true;
      }
    }
    for (int i = 0; i <= brk; i++) {
      tmpa = tot / 2;
      tmpb = (tot + 1) / 2;
      int brka = i, brkb = brk - i;
      if (brkb > tmpa || brka > tmpb) continue;
      tmpa = tmpa + i - brkb;
      tmpb = tmpb + brkb - i;
      if (tmpa < 0 || tmpb < 0) continue;
      if (a == tmpa && b == tmpb) {
        return true;
      }
    }
  } else {
    for (int i = 0; i <= brk; i++) {
      int tmpa = (tot) / 2;
      int tmpb = (tot) / 2;
      int brka = i, brkb = brk - i;
      if (brka > tmpb || brkb > tmpa) continue;
      tmpa = tmpa + i - brkb;
      tmpb = tmpb + brkb - i;
      if (tmpa < 0 || tmpb < 0) continue;
      if (a == tmpa && b == tmpb) return true;
    }
  }
  return false;
}
void solve1() {
  int a, b;
  cin >> a >> b;
  int tot = a + b;
  int idx = 0;
  for (int i = 0; i <= tot; i++) {
    if (check(i, a, b)) arr[idx++] = i;
  }
  cout << idx << endl;
  for (int i = 0; i < idx; i++) cout << arr[i] << " ";
  cout << endl;
}
void solve2() {
  int a, b;
  cin >> a >> b;
  int tot = a + b;
  int d = abs(a - b) / 2;
  if (tot & 1) {
    int idx = 0;
    for (int i = d; i <= tot - d; i++) {
      arr[idx++] = i;
    }
    cout << idx << endl;
    for (int i = 0; i < idx; i++) cout << arr[i] << " ";
    cout << endl;
  } else {
    int idx = 0;
    for (int i = d; i <= tot - d; i += 2) arr[idx++] = i;
    cout << idx << endl;
    for (int i = 0; i < idx; i++) cout << arr[i] << " ";
    cout << endl;
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve2();
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const double PI = acos(-1.0);
const double EXP = exp(1.0);
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long lowbit(long long x) { return x & (-x); }
stack<int> st;
int arr[maxn];
bool check(int brk, int a, int b) {
  int tot = a + b;
  if (tot & 1) {
    int tmpa = (tot + 1) / 2;
    int tmpb = (tot) / 2;
    for (int i = 0; i <= brk; i++) {
      tmpa = (tot + 1) / 2;
      tmpb = (tot) / 2;
      int brka = i, brkb = brk - i;
      if (brka > tmpb || brkb > tmpa) continue;
      tmpa = tmpa + i - brkb;
      tmpb = tmpb + brkb - i;
      if (tmpa < 0 || tmpb < 0) continue;
      if (a == tmpa && b == tmpb) {
        return true;
      }
    }
    for (int i = 0; i <= brk; i++) {
      tmpa = tot / 2;
      tmpb = (tot + 1) / 2;
      int brka = i, brkb = brk - i;
      if (brkb > tmpa || brka > tmpb) continue;
      tmpa = tmpa + i - brkb;
      tmpb = tmpb + brkb - i;
      if (tmpa < 0 || tmpb < 0) continue;
      if (a == tmpa && b == tmpb) {
        return true;
      }
    }
  } else {
    for (int i = 0; i <= brk; i++) {
      int tmpa = (tot) / 2;
      int tmpb = (tot) / 2;
      int brka = i, brkb = brk - i;
      if (brka > tmpb || brkb > tmpa) continue;
      tmpa = tmpa + i - brkb;
      tmpb = tmpb + brkb - i;
      if (tmpa < 0 || tmpb < 0) continue;
      if (a == tmpa && b == tmpb) return true;
    }
  }
  return false;
}
void solve1() {
  int a, b;
  cin >> a >> b;
  int tot = a + b;
  int idx = 0;
  for (int i = 0; i <= tot; i++) {
    if (check(i, a, b)) arr[idx++] = i;
  }
  cout << idx << endl;
  for (int i = 0; i < idx; i++) cout << arr[i] << " ";
  cout << endl;
}
void solve2() {
  int a, b;
  cin >> a >> b;
  int tot = a + b;
  int d = abs(a - b) / 2;
  if (tot & 1) {
    int idx = 0;
    for (int i = d; i <= tot - d; i++) {
      arr[idx++] = i;
    }
    cout << idx << endl;
    for (int i = 0; i < idx; i++) cout << arr[i] << " ";
    cout << endl;
  } else {
    int idx = 0;
    for (int i = d; i <= tot - d; i += 2) arr[idx++] = i;
    cout << idx << endl;
    for (int i = 0; i < idx; i++) cout << arr[i] << " ";
    cout << endl;
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve2();
  }
  return 0;
}

```