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;

void solve(){
    int n,k;
    cin >> n >> k;
    
    vector<int> v(n);
    
    if(k>=(n+1)/2){
        cout << -1 << endl;
        return;
    }
    
    int a1=1;
    int a2=n;
    
    for(int i=0;i<n;i++){
        if(k!=0 && i!=0 && i%2!=0){
                v[i] = a2;
                a2--;
                k--;
        }
        else{
            v[i] = a1;
            a1++;
        }
    }
    
    for(int i=0;i<n;i++){
        cout << v[i] << " ";
    }
    cout << endl;
}

int main(){
    int t;
    cin >> t;
    while(t--){
        solve();
    }
}