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())

def make(k):
    maxi = 2*k+1
    ans = [0]*(maxi)
    
    for i in range(k):
        ans[i*2+1]=maxi-k+i+1 
    for i in range(maxi-k):
        ans[i*2]= i+1
    #print('ha',ans)
    return ans
def solve(n,k):
    mini_length = k*2+1
    if n<mini_length:
        return -1
    else:
        return make(k)+[i for i in range(mini_length+1,n+1)]

for i in range(T):
    lista = list(map(int,input().split()))
    n = lista[0]
    k = lista[1]
    ans = solve(n,k)
    if ans==-1:
        print(ans)
    else:
        texto = ''
        #print(ans)
        for i in ans:
            texto+=str(i)+' '
        print(texto[:-1])