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<bits/stdc++.h>
using namespace std;
#define FastIO ios_base::sync_with_stdio(false); cin.tie(0);
#define ll long long
#define ull unsigned long long
#define pb push_back
#define All(x) (x).begin(),(x).end()
#define mp make_pair
#define nl "\n"
typedef pair<int,int>ii;
typedef vector<int>vi;
typedef vector<ii>vii;
typedef vector<vi>vvi;
typedef vector<long long>vl;
typedef vector <vl>vvl;
template<class T>
void print(T& a)
{
    for(auto x:a)
        cout<<x<<" ";
    cout<<"\n";
}
int main()
{

    FastIO

    int tt;
    cin>>tt;

    while(tt--)
    {
        int n,k,t,t1;
        cin>>n>>k;
        t=k;
        t1=n;

        if(k>(n-1)/2)
        {
            cout<<-1<<nl;
            continue;
        }

        vi a(n);
        for(int i=1; i<n-1; i+=2)
        {
            if(t==0) break;
            a[i]=t1;
            t1--;
            t--;
        }

        t=1;
        for(int i=0; i<n; i++)
        {
            if(a[i]!=0) continue;
            a[i]=t;
            t++;
        }
        print(a);
    }

    return 0;
}