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;
#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define debug(a) cerr<<#a<<" --> "<<(a)<<endl;
#define int long long
#define pii make_pair
#define pb push_back
#define ff first
#define ss second
#define all(x) x.begin(), x.end()
#define endl "\n"
int ceil(int a, int b) { if(a%b==0) return a/b; else return a/b+1; }
bool checkbit(int pos, int mask) { return mask&(1LL<<pos); }
int turnon(int pos, int mask) { return mask|(1LL<<pos); }
const int N=1005;
//int ar[N];

main()
{
    fastio;
    int t;
    cin>>t;
    while(t--) {
        int a, b;
        set<int>st;
        cin>>a>>b;
        int tot=a+b;
        int serve=ceil(tot, 2);//alice serve
        int bs=tot-serve;
        for(int i=serve; i>=0; i--) {
            if(a<serve) {
                //debug(i-a);
                int baki=serve-i;
                st.insert(i-a+baki);
            }
            else {
                int baki=a-i;
                if(baki>bs) break;
                int bakis=serve-i;
                //debug(baki+bakis);
                st.insert(baki+bakis);
            }
        }
        serve=tot/2;
        bs=tot-serve;
        for(int i=serve; i>=0; i--) {
            if(a<serve) {
                //debug(i-a);
                int baki=serve-i;
                st.insert(i-a+baki);
            }
            else {
                int baki=a-i;
                if(baki>bs) break;
                int bakis=serve-i;
                //debug(baki+bakis);
                st.insert(baki+bakis);
            }
        }
        cout<<st.size()<<endl;
        for(auto i:st) cout<<i<<" ";
        cout<<endl;
    }
}