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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {

    void run() throws IOException {
        int n = nextInt();
        int k = nextInt();
        if (k > (n - 1) / 2) out.println(-1);
        else {
            int l = 1;
            int r = n;
            int z = 0;
            while (k > 0) {
                out.print(l + " ");
                out.print(r + " ");
                l++;
                r--;
                k--;
                z += 2;
            }
            while (z < n) {
                out.print(l + " ");
                z++;
                l++;
            }
            out.println();
        }
    }

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out = new PrintWriter(System.out);
    StringTokenizer in = new StringTokenizer("");


    // pairs
    class pil implements Comparable<pil> {
        int first;
        long second;

        pil(int f, long s) {
            this.first = f;
            this.second = s;
        }

        public int compareTo(pil o) {
            if (first != o.first) return Integer.compare(first, o.first);
            return Long.compare(second, o.second);
        }
    }

    class pii implements Comparable<pii> {
        int fi;
        int se;

        pii(int s, int f) {
            se = f;
            fi = s;
        }

        public int compareTo(pii o) {
            if (fi != o.fi) return Integer.compare(fi, o.fi);
            return Integer.compare(se, o.se);
        }
    }

    class vert {
        int to;
        int iter;
        int idx;
        int ed;

        vert(int s, int f, int zz, int gg) {
            to = s;
            iter = f;
            idx = zz;
            ed = gg;
        }
    }


    class pll implements Comparable<pll> {
        long first;
        long second;

        pll(long f, long s) {
            this.first = f;
            this.second = s;
        }

        public int compareTo(pll o) {
            if (first != o.first) return Long.compare(first, o.first);
            return Long.compare(second, o.second);
        }
    }

    class pli implements Comparable<pli> {
        long first;
        int second;

        pli(long f, int s) {
            this.first = f;
            this.second = s;
        }

        public int compareTo(pli o) {
            if (first != o.first) return Long.compare(first, o.first);
            return Integer.compare(second, o.second);
        }
    }

    boolean hasNext() throws IOException {
        if (in.hasMoreTokens()) return true;
        String s;
        while ((s = br.readLine()) != null) {
            in = new StringTokenizer(s);
            if (in.hasMoreTokens()) return true;
        }
        return false;
    }

    String nextToken() throws IOException {
        while (!in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public static void main(String[] args) throws IOException {
        Main m = new Main();
        int t = m.nextInt();
        while (t-- > 0)
            m.run();
        m.out.close();
    }
}