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 sys import stdin as si , stdout as so

for _ in range(int(si.readline())):
    n , k = map(int , si.readline().split())
    if k==0:
        for i in range(1 , n+1):
            so.write(str(i) + " ")
        so.write("\n")
    elif (n==2 and k>0) or n==k or(n%2==0 and k==n//2) or k>n//2:
        so.write(str(-1) + "\n")
    else:
        c = 0
        d = n
        e = 1
        for i in range(1 , n+1):
            if i%2!=0:
                so.write(str(e) + " ")
                e += 1
            elif c < k:
                so.write(str(d) + " ")
                d -= 1
                c += 1
            else:
                so.write(str(e) + " ")
                e += 1
        so.write("\n")