Problem: You are given two integers n and k. You are asked to choose maximum number of distinct integers from 1 to n so that there is no subset of chosen numbers with sum equal to k.

A subset of a set is a set that can be obtained from initial one by removing some (possibly all or none) elements of it.

Input

The first line contains the number of test cases T (1 ≤ T ≤ 100).

Each of the next T lines contains two integers n and k (1 ≤ k ≤ n ≤ 1000) — the description of test cases.

Output

For each test case output two lines. In the first line output a single integer m — the number of chosen integers.

In the second line output m distinct integers from 1 to n — the chosen numbers.

If there are multiple answers, print any. You can print the numbers in any order.

Example

Input


3
3 2
5 3
1 1


Output


2
3 1 
3
4 5 2 
0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
        
using namespace std;
    
#define ar array
#define ll long long
    
const int MAX_N = 1e5 + 1;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;

void solve() {
    
    ll n, k;
    cin>>n>>k;

    ll p = k/2;
    ll ans = p + n-k;

    if(k >= (2*n)) {
        cout<<n<<endl;
        for(ll i=1;i<n;i++) {
            cout<<i<<" ";
        }
        cout<<n<<endl;
        return;
    }

    if(n == 1) {
        cout<<"0"<<endl;
        return;
    }

    vector<ll> v;
    for(ll i=ceil(k/2.);i<min(n+1,k);i++) {
        v.push_back(i);
    }

    for(ll i=k+1;i<=n;i++) {
        v.push_back(i);
    }

    cout<<v.size()<<endl;
    for(ll i=0;i<v.size()-1;i++) {
        cout<<v[i]<<" ";
    }
    cout<<v[v.size()-1]<<endl;

}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    int tc = 1;
    cin >> tc;
    for (int t = 1; t <= tc; t++) {
        // cout << "Case #" << t  << ": ";
        solve();
    }
}