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;
const int maxn = 5e5 + 5;
int s[maxn];
void solve() {
  long long n, m, i, j, f = 1, k, b, a, t = 0, c;
  scanf("%lld%lld", &a, &b);
  n = a + b;
  a = min(a, b);
  c = n / 2;
  if (n % 2) {
    for (i = n - (c + 1 + a); i <= c + 1; i += 2) {
      s[++t] = i;
      if (t == a + 1) break;
    }
    if (s[t] == c + 1) s[t]--;
    printf("%d\n", t * 2);
    for (i = 1; i <= t; i++) printf("%d ", s[i]);
    for (i = t; i >= 1; i--) printf("%d ", n - s[i]);
    printf("\n");
  } else {
    for (i = n - (c + a); i <= c; i += 2) {
      s[++t] = i;
    }
    if (s[t] == c) {
      printf("%d\n", t * 2 - 1);
      for (i = 1; i < t; i++) printf("%d ", s[i]);
      for (i = t; i >= 1; i--) printf("%d ", n - s[i]);
    } else {
      printf("%d\n", t * 2);
      for (i = 1; i <= t; i++) printf("%d ", s[i]);
      for (i = t; i >= 1; i--) printf("%d ", n - s[i]);
    }
    printf("\n");
  }
}
int main() {
  int T = 1;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}
