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;
using ll = long long;

void solve() {
    int n, k;
    cin >> n >> k;
    if(2 * k + 1 > n) {
        cout << "-1\n";
        return;
    }
    vector<int> v(n);
    int p = n;
    for(int i = 1; i < 1 + 2 * k; i += 2)
        v[i] = p--;
    p = 1;
    for(int i = 0; i < n; i++) {
        if(v[i])
            continue;
        v[i] = p++;
    }
    for(auto &x : v)
        cout << x << " ";
    cout << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);

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