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 PYTHON3):

Solution:
from os import path
import sys,time
from math import ceil, floor,gcd,log,log2 ,factorial
from collections import defaultdict ,Counter , OrderedDict , deque
from heapq import heapify , heappush , heappop
from bisect import *
# from functools import reduce
from operator import mul
maxx, mod  = float('inf') , int(1e9 + 7)
localsys ,start_time = 0 , time.time()
if (path.exists('input.txt')):
    localsys = 1;sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
#left shift --- num*(2**k) --(k - shift)
input = sys.stdin.readline
for _ in range(int (input())):
    #3 , 5 , 7
    n , k = map(int , input().split())
    if n < 3 and k >0:
        print(-1)
        continue
    if k == 0:
        print(*list(range(1, n+1)))
        continue
    x = (n-3)//2 + 1
    if k > x:
        print(-1)
        continue
    ans =[]
    i = 1 ; j = n
    cnt =0
    el = 3 + (k-1)*2
    for t in range(n):
        if cnt < el:
            if cnt % 2 ==0:
                ans.append(i)
                i+=1
            else:
                ans.append(j)
                j-=1
        else:
            ans.append(i)
            i+=1
        cnt+=1
    print(*ans)
















if localsys:
    print("\n\n\nTime Elased :",time.time() - start_time,"seconds")