#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
long long fxp(long long a, long long b) {
  assert(b >= 0);
  a %= mod;
  long long ans = 1;
  while (b > 0) {
    if (b % 2) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return ans % mod;
}
inline long long minv(long long x) { return fxp(x, mod - 2); }
void c_p_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int mXn = 2e5 + 4;
const long long inf = 1e18;
vector<int> ans;
int ck = 0;
void brute(int lev, int arem, int brem) {
  if (arem + brem == 0) {
    ans.push_back(ck);
    return;
  }
  if (arem) {
    if (lev == 1) ck++;
    brute(lev ^ 1, arem - 1, brem);
    if (lev == 1) ck--;
  }
  if (brem) {
    if (lev == 0) ck++;
    brute(lev ^ 1, arem, brem - 1);
    if (lev == 0) ck--;
  }
}
signed main() {
  c_p_c();
  int T;
  cin >> T;
  for (long long tc = 1; tc <= T; tc++) {
    int a, b;
    cin >> a >> b;
    int s = (a + b);
    pair<int, int> x[2];
    x[0] = {s / 2, s - s / 2};
    x[1] = {s - s / 2, s / 2};
    int mn = max(0, a - x[0].first) + max(0, b - x[0].second);
    mn = min(mn, max(0, a - x[1].first) + max(0, b - x[1].second));
    int mx = min(a, x[0].second) + min(b, x[0].first);
    mx = max(mx, min(a, x[1].second) + min(b, x[1].first));
    vector<int> bb;
    if (s % 2) {
      for (long long i = mn; i < mx + 1; i++) bb.push_back(i);
    } else {
      for (int i = mn; i <= mx; i += 2) bb.push_back(i);
    }
    cout << bb.size() << '\n';
    for (auto x : bb) cout << x << ' ';
    cout << '\n';
  }
}

### Prompt

Your challenge is to write 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;
const int mod = 1e9 + 7;
long long fxp(long long a, long long b) {
  assert(b >= 0);
  a %= mod;
  long long ans = 1;
  while (b > 0) {
    if (b % 2) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return ans % mod;
}
inline long long minv(long long x) { return fxp(x, mod - 2); }
void c_p_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int mXn = 2e5 + 4;
const long long inf = 1e18;
vector<int> ans;
int ck = 0;
void brute(int lev, int arem, int brem) {
  if (arem + brem == 0) {
    ans.push_back(ck);
    return;
  }
  if (arem) {
    if (lev == 1) ck++;
    brute(lev ^ 1, arem - 1, brem);
    if (lev == 1) ck--;
  }
  if (brem) {
    if (lev == 0) ck++;
    brute(lev ^ 1, arem, brem - 1);
    if (lev == 0) ck--;
  }
}
signed main() {
  c_p_c();
  int T;
  cin >> T;
  for (long long tc = 1; tc <= T; tc++) {
    int a, b;
    cin >> a >> b;
    int s = (a + b);
    pair<int, int> x[2];
    x[0] = {s / 2, s - s / 2};
    x[1] = {s - s / 2, s / 2};
    int mn = max(0, a - x[0].first) + max(0, b - x[0].second);
    mn = min(mn, max(0, a - x[1].first) + max(0, b - x[1].second));
    int mx = min(a, x[0].second) + min(b, x[0].first);
    mx = max(mx, min(a, x[1].second) + min(b, x[1].first));
    vector<int> bb;
    if (s % 2) {
      for (long long i = mn; i < mx + 1; i++) bb.push_back(i);
    } else {
      for (int i = mn; i <= mx; i += 2) bb.push_back(i);
    }
    cout << bb.size() << '\n';
    for (auto x : bb) cout << x << ' ';
    cout << '\n';
  }
}

```