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 endl '\n'
long long nt,k,n,curr;
long long a[107];

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    //freopen("input.txt", "r", stdin);
    cin >> nt;
    for(int t=1; t<=nt; t++) {
        cin >> n >> k;
        for(int i=1; i<=n; i++) a[i]=0;
        if(2*k+1>n) {
            cout << -1<<endl;
            continue;
        }
        for(int i=1; i<=k; i++) a[2*i] = n+1-i;
        curr = n-k;
        for(int i=1; i<=n; i++)
            if(a[i]==0) {
                a[i]=curr;
                curr--;
            }
        for(int i=1; i<=n; i++) cout << a[i]<<" ";
        cout << endl;
    }

    return 0;
}