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 n714A {

    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int t = sc.ni();
        while(t-- > 0) {
            int n = sc.ni();
            int k = sc.ni();
            if ((n - 1) / 2 < k)
                System.out.println(-1);
            else {
                if (n < 3 & k > 0)
                    System.out.println(-1);
                else {
                    int i = 1;
                    while (i <= n) {
                        if (k == 0) {
                            System.out.print(i + " ");
                            i++;
                        } else {
                            if (i == 1) {
                                System.out.print(1 + " " + 3 + " " + 2 + " ");
                                i += 3;
                                k--;
                            } else {
                                if (i + 1 <= n) {
                                    System.out.print((i + 1) + " " + (i) + " ");
                                    i += 2;
                                    k--;
                                } else {
                                    System.out.print(i);
                                }
                            }
                        }
                    }
                    System.out.println();
                }
            }
        }
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in), 32768);
            st = null;
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int ni() {
            return Integer.parseInt(next());
        }

        int[][] graph(int N, int[][] edges) {
            int[][] graph = new int[N][];
            int[] sz = new int[N];
            for (int[] e: edges) {
                sz[e[0]] += 1;
                sz[e[1]] += 1;
            }
            for (int i = 0; i < N; i++) {
                graph[i] = new int[sz[i]];
            }
            int[] cur = new int[N];
            for (int[] e: edges) {
                graph[e[0]][cur[e[0]]] = e[1];
                graph[e[1]][cur[e[1]]] = e[0];
                cur[e[0]] += 1;
                cur[e[1]] += 1;
            }
            return graph;
        }

        int[] intArray(int N, int mod) {
            int[] ret = new int[N];
            for (int i = 0; i < N; i++)
                ret[i] = ni()+mod;
            return ret;
        }

        long nl() {
            return Long.parseLong(next());
        }

        long[] longArray(int N, long mod) {
            long[] ret = new long[N];
            for (int i = 0; i < N; i++)
                ret[i] = nl()+mod;
            return ret;
        }

        double nd() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}