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. 
#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
constexpr long long pct(long long x) { return __builtin_popcount(x); }
const long long DFS_WHITE = 0;
const long long DFS_BLACK = 1;
const long long INF = 1e18;
const long long MAXN = 2e5 + 10;
const long long MOD = 1e9 + 7;
void setIO(string filename = "") {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  if (filename.size() == 0) return;
  freopen((filename + ".in").c_str(), "r", stdin);
  freopen((filename + ".out").c_str(), "w", stdout);
}
long long n, m;
void solve() {
  cin >> n >> m;
  if (m > n) {
    swap(n, m);
  }
  if ((m + n) % 2 == 0) {
    long long tmp = (n + m) / 2;
    cout << (n - tmp + m * 2 - (n - tmp)) << endl;
    for (long long i = n - tmp; i <= n - tmp + m * 2; i += 2) {
      cout << i << " ";
    }
  } else {
    vector<long long> ans;
    long long tmp = (n + m + 1) / 2;
    for (long long i = n - tmp; i <= n - tmp + m * 2; i += 2) {
      ans.push_back(i);
    }
    tmp = (n + m - 1) / 2;
    for (long long i = n - tmp; i <= n - tmp + 2 * m; i += 2) {
      ans.push_back(i);
    }
    vector<long long> res;
    long long cur = 0;
    res.push_back(ans[cur]);
    for (long long i = 0; i < ans.size(); i++) {
      if (ans[i] != ans[cur]) {
        cur++;
        res.push_back(ans[i]);
      }
    }
    sort(res.begin(), res.end());
    cout << res.size() << endl;
    for (long long i = 0; i < res.size(); i++) {
      cout << res[i] << " ";
    }
  }
}
signed main() {
  setIO();
  long long test = 1;
  cin >> test;
  while (test--) {
    solve();
    if (test) cout << endl;
  }
}
