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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class A {

    public static void main(String[] args) {
        FastScanner fs=new FastScanner();
        int T=fs.nextInt();
        for (int tt=1; tt<=T; tt++) {
            int n = fs.nextInt();
            int k = fs.nextInt();
            int maxP = (n / 2) + (n % 2) - 1;
            if(k <= maxP) {
                int[] p = new int[n];
                p[0] = 1; int st = 2;
                for(int i = 1; i < n; i++) {
                    if(k > 0) {
                        p[i + 1] = st++;
                        p[i] = st++; k--; i++;
                    } else p[i] = st++;
                }
                for(int i = 0; i < n; i++) {
                    System.out.print(p[i]);
                    if(i < n - 1) System.out.print(" ");
                }
                System.out.println();
            } else System.out.println(-1);
        }
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long[] readLongArray(int n) {
            long[] a=new long[n];
            for (int i=0; i<n; i++) a[i]=nextLong();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }
}