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
#include <bits/stdc++.h>
using namespace std;
//#define int long long int
#define fast                 \
    ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0);
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
#define F first
#define S second
#define PB push_back
#define MP make_pair
const ll mod = 1e9 + 7, N = 2e6 + 7, M = 1e6 + 7, INF = INT_MAX / 10;

/*void solve()
{
    int n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    int str[26];
    int ct1[26],ct2[26];
    for (int i = 0; i < 26;i++){
        ct1[i] = 0;
        ct2[i] = 0;
        str[i] = n;
    }
    if (n % k)
        {
            cout << -1 << endl;
            return;
        }
    for (int i = 0; i < s.length(); i++)
    {
        ct1[s[i] - 'a']++;
        ct2[s[i] - 'a']++;

    }
    vector<int> v;
    int idx = n;
    for (int i = 0; i <s.length(); i++)
    {
        if(ct2[s[i]-'a']/k>0)
            continue;
        else if((ct2[s[i]-'a']/k)==0 && (ct1[s[i]-'a']%k)){
            v.PB(i);
        }
        else if(v.size()>0 && (ct2[s[i]-'a']/k)==1 && (ct1[s[i]-'a']%k)==0){

        }
        ct2[i]--;
    }
    if(idx==n){
        cout << s << endl;
        return;
    }
    else{
        sort(v.begin(),v.end());
        int ans = 0;
        string res = s;
        int f = k;
        res[v[0]] = (char)(s[v[0]] - 'a' + (int)('a'));
        if (s[v[0]] - 'a' + 1 < 26)
        {
           if(f){
            for (int i = v.size() - 1; i > idx; i--)
            {
                res[v[i]] = (char)(s[v[0]] - 'a' + (int)('a'));
            }
            f--;
           }
           else{
               res[v[i]] = 'a';
           }
        }
        cout << res << endl;
        for (int i = s[v[0]] - 'a'; i < 26; i++)
        {
            for (int i = 0; i < v.size(); i++)
            {
                if()
            }
        }
    }
}*/
void solve()
{
    int n, k;
    cin >> n >> k;
    int f = k / 2;
    int ans = n - k;
    ans += k / 2;
    if (k % 2)
        f++;
    cout << ans << endl;
    for (int i = f; i <= n; i++)
    {
        if (i != k)
            cout << i << " ";
    }
    cout << endl;
}
signed main()
{
    fast;
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}
