Problem: A sequence of n integers is called a permutation if it contains all integers from 1 to n exactly once.

Given two integers n and k, construct a permutation a of numbers from 1 to n which has exactly k peaks. An index i of an array a of size n is said to be a peak if 1 < i < n and a_i \gt a_{i-1} and a_i \gt a_{i+1}. If such permutation is not possible, then print -1.

Input

The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases.

Then t lines follow, each containing two space-separated integers n (1 ≤ n ≤ 100) and k (0 ≤ k ≤ n) — the length of an array and the required number of peaks.

Output

Output t lines. For each test case, if there is no permutation with given length and number of peaks, then print -1. Otherwise print a line containing n space-separated integers which forms a permutation of numbers from 1 to n and contains exactly k peaks. 

If there are multiple answers, print any.

Example

Input


5
1 0
5 2
6 6
2 1
6 1


Output


1 
2 4 1 5 3 
-1
-1
1 3 6 5 4 2

Note

In the second test case of the example, we have array a = [2,4,1,5,3]. Here, indices i=2 and i=4 are the peaks of the array. This is because (a_{2} \gt a_{1} , a_{2} \gt a_{3}) and (a_{4} \gt a_{3}, a_{4} \gt a_{5}). 

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#include <string>
#include <cmath>
using namespace std;

#define fio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define ll long long
#define vi vector<int>
#define pii pair<int,int>
#define pb push_back
#define fi first
#define se second
#define all(v) v.begin(),v.end()
#define printV(v) for(int i = 0; i < v.size(); i++)cout<<v[i]<<"\n"<<" \n"[i == v.size()-1]
#define printA(a) int N=sizeof(a)/sizeof(a[0]);for(int i = 0; i < N; i++)cout<<a[i]<<" "<<" \n"[i==N-1]

int main()
{
    fio
    int T;
    cin >> T;
    while(T--)
    {
        int n, k;
        cin >> n >> k;
        int i = 1, j = n;
        vi v;
        int mx = ((n+1)/2) - 1;
        if(k==0)
        {
            for(int i=1; i<=n; i++)
                cout<<i<<" ";
            cout<<"\n";
            continue;
        }
        while(i < j)
        {
            v.pb(i);
            v.pb(j);
            i = i+1;
            j--;
            //cout<<"#"<<i<<" "<<j<<endl;
        }
        if(n&1)
            v.pb(j);
        //cout << i << " " << j << endl;
        int cnt = 0, ind = -1;
        vi x;
        for(int i=1; i<v.size()-1; i++)
        {
            if(cnt==k)
            {
                ind = i+1;
                for(int j=i+1; j<v.size(); j++)
                    x.pb(v[j]);
                break;
            }
            if(v[i] > v[i-1] && v[i] > v[i+1])
            {
                cnt++;
            }
        }
        //cout<<"=>"<< ind<<" "<< cnt <<endl;
        if(cnt==k && k == mx)
        {
            for(auto i: v)
                cout<<i<<" ";
            cout<<"\n";
            continue;
        }
        if(ind==-1)
        {
            cout<<ind<<"\n";
            continue;
        }
        sort(all(x));
        for(int i=0; i<ind; i++)
            cout<<v[i]<<" ";
        for(auto i: x)
            cout<<i<<" ";
        cout<<"\n";
    }
    return 0;
}