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.StringTokenizer;

public class A {

	public static void main(String[] args) {
		FastScanner fs=new FastScanner();
		int T=fs.nextInt();
		for (int tt=0; tt<T; tt++) {
			int n=fs.nextInt(), k=fs.nextInt();
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=i+1;
			
			if (k*2>=n) {
				System.out.println(-1);
				continue;
			}
			for (int i=0; i<k; i++) {
				int i1=i*2+1, i2=i1+1;
				int temp=a[i1];
				a[i1]=a[i2];
				a[i2]=temp;
			}
			for (int i:a) System.out.print(i+" ");
			System.out.println();
		}
		
	}
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		public String next() {
			while (!st.hasMoreElements())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int[] readArray(int n) {
			int[] a=new int[n]; 
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		int nextInt () {
			return Integer.parseInt(next());
		}
	}

}