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 {
	static FastReader sc=null;
	
	public static void main(String[] args) {
		sc=new FastReader();
		PrintWriter pw=new PrintWriter(System.out);
		int t=sc.nextInt();
		for(int o=1;o<=t;++o) {
			int a=sc.nextInt();
			int n=sc.nextInt();
			int ans[]=new int[a];
			if(n>(a-1)/2) {
				System.out.println("-1");
				continue;
			}
			if(a==1) {
				System.out.println("1");
				continue;
			}
			int q=0;
			for(int i=1;i<a;i+=2,q++) {
				if(q==n)break;
				ans[i]=a-q;
			}
			q=1;
			for(int i=0;i<a;i++,q++) {
				if(ans[i]==0)ans[i]=q;
				else q--;
			}
			for(int i:ans) {
				System.out.print(i+" ");
			}
			System.out.print("\n");
		}
		pw.close();
	}
/*                                                   中																*/
	static void put(int a[]) {
		for(int i:a) {
			System.out.print(i+" ");
		}
		System.out.println("\n");
	}
	static void copy(int a[],int b[]) {
		System.arraycopy(a, 0, b, 0, a.length);
		return;
	}
	static long gcd(long a,long b) {
		if(b==0)return a;
		else return gcd(b,a%b);
	}
	static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
        int[] readArray(int n) {
    		int a[]=new int [n];
    		for(int i=0;i<n;i++) {
    			a[i]=sc.nextInt();
    		}
    		return a;
    	}
    } 
}
/*Arrays.sort(a,new Comparator<Integer>(){
				public int compare(Integer o1, Integer o2) {
					if(o1 < o2){
	                    return 1;
	                }else if(o1 > o2){
	                    return -1;
	                }
					return 0;
				}
			}
			);必须用Integer声明数组*/