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 math
for _ in range(int(input())):
    n,k=map(int,input().split())
    t=0
    if n%2==0:
        t=math.floor((n-1)/2)
    else:
        t=math.floor(n/2)
    if k>t or (n<=2 and k!=0):
        print("-1")
    else:
        newlist = [x for x in range(1,n+1)]
        temp=0
        for i in range(1,n):
            if i%2==0 and k!=0:
                temp=newlist[i-1]
                newlist[i-1]=newlist[i]
                newlist[i]=temp
                k-=1
                
        print(*newlist)