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,math
inf = float('inf')
input = sys.stdin.readline
def lcm(a,b):
    return int((a/math.gcd(a,b))*b)
def gcd(a,b):
    return int(math.gcd(a,b))
"""
n = int(input())
n,k = map(int,input().split())
arr = list(map(int,input().split()))
"""
def checkpeak(arr):
    c = 0
    for i in range(1,len(arr)-1):
        if arr[i]==0:
            return c
        if arr[i]>arr[i-1] and arr[i]>arr[i+1]:
            c+=1
    return c

for _ in range(int(input())):
    n,k = map(int,input().split())
    if n==1:
        if k==1:
            print(-1)
        else:
            print(1)
    else:
        ans       =  [0]*n
        avail     =  [i for i in range(1,n+1)]
        peakCount =  0
        for i in range(n):
            if i%2==0:
                ans[i]=avail.pop(0)
            else:
                if checkpeak(ans)<k:
                    ans[i]=avail.pop()
                else:
                    ans[i]=avail.pop(0)

        if checkpeak(ans)==k:
            print(*ans)
        else:
            print(-1)