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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;

public class CFArraysAndPeaks {
	
	public static void main(String args[]) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
		
		
		String T = br.readLine();
		int t = Integer.parseInt(T);
		while(t-- >0) {
			String inp[] = br.readLine().split(" ");
			
			int n = Integer.parseInt(inp[0]);
			int k = Integer.parseInt(inp[1]);
			
			int arr[] = new int[n];
			int smallest=1;
			int largest=n;
			int indx=1;
			if(n==1) {
				if(k==0) {bw.write(n+"\n");}
				else {bw.write(-1+"\n");}
				continue;
			}
			if(2*k+1 >n) {
				bw.write(-1+"\n");
				continue;
			}
			int kCopy=k;
			for(int i=1;i<=kCopy;i++) {
				arr[indx]=largest-k+1;
				k--;
				indx=indx+2;
			}
			for(int i=0;i<n;i++) {
				if(arr[i]==0) {
					arr[i]=smallest;
					smallest++;
				}
			}
			for(int i=0;i<n;i++) {
				bw.write(arr[i]+" ");
			}
			bw.write("\n");
		}bw.flush();
		
	}
}