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>
#include <cmath>
#include <sstream>
#include <vector>
#include <stdio.h>
#define ll              long long
#define fr(i,j,p)       for(ll i=j; i<p ;i++)
#define frr(i,j,p)      for(int i=j; i>=p ;i--)
#define PI              3.14159265358979323846
#define print(m)        std::cout << std::fixed << std::setprecision(11) << m <<endl;
#define mkp             make_pair
#define pb              push_back
#define M               32
#define lbindex(a,x)    (int)(lower_bound(a.begin(), a.end(),x)-a.begin());
#define upindex(a,x)    (int)(upper_bound(a.begin(), a.end(),x)-a.begin());
#define all(a)          (a).begin(),(a).end()
#define mem(b,a,n)      memset(b,a,sizeof(b[0])*n);
#define meme(b,a,n,m)   memset(b,a,sizeof(b[0][0])*n*m);
#define ff              first
#define ss              second
#define endl            '\n'

//ll md=1000000007;
//ll mdd=998244353;

using namespace std;

typedef vector<ll> vl;		typedef vector<int> vi;
typedef vector<vl> vvl;	  	typedef vector<vi> vvi;
typedef pair<ll, ll> pll;	typedef pair<int, int> pii;

int main()
{
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(NULL);  cout.tie(NULL);
    int t1=1;       cin>>t1;
    while(t1--)
    {
        int n,k;    cin>>n>>k;
        if(k>(n-1)/2)   {cout<<-1<<endl;    continue;}
        vi a(n, 0);
        int x=n;
        for(int i=1;i<n-1;i+=2)
        {
            if(k==0)    break;
            a[i]=x; x--;    k--;
        }
        fr(i,0,n)
        {
            if(a[i]==0) {a[i]=x; x--;}
        }
        fr(i,0,n)   cout<<a[i]<<' ';
        cout<<endl;
    }
    return 0;
}