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.util.*;
import java.io.*;

public class Rextester {
    static final FS sc          = new FS();
    static final PrintWriter pw = new PrintWriter(System.out);
    static final Util u         = new Util();
    
    public static void main(String[] args) {
        int testCases = sc.nextInt();
        while(testCases-- > 0) {
            int n = sc.nextInt();
            int peaks = sc.nextInt();
            
            int maxPeaksPossible = (n-1)/2;
            if(maxPeaksPossible < peaks)
                pw.println(-1);
            else {
                int[] ans = new int[100];
                
                int val = 1;
                int i;
                for(i = 0; i<(2*peaks+1); i+=2) ans[i] = val++;
                for(i = 1; i<(2*peaks+1); i+=2) ans[i] = val++;
                for(; i<n; i++) ans[i] = val++;
                
                for(int v : ans) {
                    if(v != 0)
                        pw.print(v+" ");
                } pw.println();
            }
        }
        pw.flush();
    }
    
    
    
    
    //------------------------------Utility-Classes----------------------------//
    static class FS {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        String next() {
            while(!st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch(Exception e) {}
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }
    static class Util {
        
        void sc1D(int[] a) {
            for(int i = 0; i<a.length; i++)
                    a[i] = sc.nextInt();
        }

        void sc2D(int[][] a) {
            for(int i = 0; i<a.length; i++)
                for(int j = 0; j<a[0].length; j++)
                        a[i][j] = sc.nextInt();
        }
        
        int gcd(int number1, int number2) {
            if (number2 == 0)
                return number1;
            return gcd(number2, number1 % number2);
        }
        
    }
}