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(){
    long long n, k;
    cin >> n >> k;
    vector<long long> v;
    long long i = 1;
    v.push_back(i);
    while(i <= n && k > 0){
        v.push_back(i + 2);
        v.push_back(i + 1);
        i += 2;
        k--;
    }
    if(k != 0 || v.size() > n) cout << "-1\n";
    else{
        for(long long i = 0 ; i < v.size() ; i++){
            cout << v[i] << " ";
        }
        long long j = v.size() + 1;
        while(j <= n) {
            cout << j << " ";
            j++;
        }
        cout << "\n";
    }
    return;
}

int main(){
    long long T;
    cin >> T;
    for(long long t = 0 ; t < T ; t++){
        solve();
    }
    return 0;
}