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:
/*****************************
      *   Mahfuz   *
   * Created: 12.04.2021 *
*****************************/

#include <bits/stdc++.h>
using namespace std;
#define PI 3.141592653589793238462
#define ll long long
#define ld long double
#define max3(a, b, c) max(c, max(a, b))
#define max4(a, b, c, d) max(d, max(c, max(a, b)))
#define min3(a, b, c) min(c, min(a, b))
#define min4(a, b, c, d) min(d, min(c, min(a, b)))
#define FIO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(0);
#define nl '\n'
#define mod 1000000007

int gcd(int a, int b)
{
    if(b==0)return a;
    return gcd(b, a%b);
}


int main()
{
    FIO;
    int t, n, k, i, j, a, b, c;

    cin >> t;

    while(t--)
    {
        cin >> n >> k;
        a = n-1;

        int d = a/2;

        if(k == 0)
        {
            for(i=1; i<=n; i++)cout << i << " ";
            cout << nl;
            continue;
        }

        if(k > d)
        {
            cout << "-1\n";
            continue;
        }

        vector<int>test;
        c = 0;

        for(i=n; i>0; i--)
        {
            if(c == k)break;
            else
            {
                test.push_back(i);
            }
            c++;
        }

        j = -1;

        for(i=1; i<=n-k; i++)
        {
            if(j+1 < test.size())cout << i << " " << test[++j] << " ";
            else cout << i << " ";
        }

        cout << nl;
    }

    return 0;
}