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;
typedef long long ll ;
#define sz(x) ((int)(x).size()) 
#define all(x) x.begin() , x.end()
#define rall(x) x.rbegin(), x.rend()
#define mem(arr,val) memset(arr,val,sizeof(arr))
#define deb(x) cout<<#x<<"="<<x<<endl
#define debs(x,y) cout<<#x<<"="<<x<<","<<#y<<"="<<y<<endl
#define ppc __builtin_popcount
#define ppcll __builtin_popcountll

const int mod = 1000000007;
const int MAX = INT_MAX;
const int MIN = INT_MIN;
const char nl = '\n';
const long long INF = 1e18; 

void solve(){
    int n , k ;  cin>>n>>k ;
    vector<int> a(n) ;
    if(2*k+1>n){
        cout<<"-1"<<nl ;
    }
    else{
        int cnt=0 , val = n  ; 
        for(int i=1;i<n;i+=2){
            if(cnt==k)break ;
            cnt++ ;
            a[i] = val ; 
            val-- ;
        }
        for(int i=0;i<n;i++){
            if(a[i]==0){
                a[i] = val;
                val-- ;
            }
        }
        for(int i=0;i<n;i++){
            cout<<a[i]<<" " ;
        }
        cout<<nl ;
    }

}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.precision(10);cout << fixed;
#ifndef ONLINE_JUDGE
    freopen("F:/input.txt", "r", stdin);
    freopen("F:/output.txt", "w", stdout);
#endif
    int T = 1;
    cin >> T;
    while (T--) {
        solve();
    } 
}