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:
import sys

n_list = []
k_list = []
idx = 0
t = 0
for line in sys.stdin:
    items = line.split()
    if idx == 0:
        t = int(items[0])
        idx = 1
        continue
    n_list.append(int(items[0]))
    k_list.append(int(items[1]))
    idx += 1
    if idx > t:
        break

for i in range(t):
    n = n_list[i]
    k = k_list[i]
    if (n - 1) / 2 < k:
        print("-1")
        continue
    print_string = ""
    start = 1
    end = n
    for i in range(k):
        print_string += str(start) + " " + str(end) + " "
        start += 1
        end -= 1
    
    for i in range(end - start + 1):
        print_string += str(start) + " "
        start += 1
    print(print_string)