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:
t=int(input())
for i in range(0,t):
    arr = list(map(int,input().strip().split()))
    n = arr[0]
    k = arr[1]
    temp = [0 for i in range(n)]
    a = [i+1 for i in range(n)]
    l = 0
    #print(a)
    if n>2*k:
        for j in range(0,n):
            if k!=0:
                if j%2==0:
                    m = min(a)
                    temp[j] = m
                    a.remove(m)
                else:
                    k = k-1
                    m = max(a)
                    temp[j] = m
                    a.remove(m)
            else:
                a.sort()
                temp[j] = a[l]
                l+=1
        print(*temp, sep=' ')
    else:
        print(-1)