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 ar array
#define ll long long
#define debug1(x) cout<<#x<<" "<<x<<endl;
#define debug2(x,y) cout<<#x<<" "<<x<<" "<<#y<<" "<<y<<endl;
#define debug3(x,y,z) cout<<#x<<" "<<x<<" "<<#y<<" "<<y<<" "<<#z<<" "<<z<<endl;
const int MAX_N = 1e5 + 1;
const int MOD = 998244353;
const int INF = 1e9;
const ll LINF = 1e18;
void solve() {
    int n, k; cin >> n >> k;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; ++i) a[i] = i;
    int p = 0;
    bool done =  false;
    for (int i = 1; i <= n; ++i) {
        if (p - 1 == k) {
            done = true;
            sort(a.begin() + i, a.end());
            break;
        }
        if (i % 2 == 1) {
            if (i != 1) a[i] = i - 1;
            else a[i] = 1;
            p++;
        }
        else {
            a[i] = i + 1;
        }
    }
    if (!done && p - 1 == k) {
        done = true;

    }
    if (done) {
        for (int i = 1; i <= n; ++i) {
            cout << a[i] << " ";
        }
        cout << endl;
    }
    else cout << -1 << endl;
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
#ifndef ONLINE_JUDGE
    freopen("input1.txt", "r", stdin);
    freopen("output1.txt", "w", stdout);
#endif
    int t = 1;
    cin >> t;
    for (int i = 0; i < t; ++i)
    {
        solve();
    }
}