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 _ in range(t):
    n,k = map(int,input().split())
    l = [i for i in range(1,n+1)]
    n1 = n
    if n1%2==1:
        if k>n1//2:
            print(-1)
    if n1%2==0:
        n1-=1
        if k>n1//2:
            print(-1)
    if k<=n//2 and k<=n1//2:
        peakL = l[n-k:]
        
        restL=  l[:n-k]
        #print(peakL, restL)
        ans = []
        k = 0
        for i in range(len(restL)):
            ans.append(restL[i])
            if k<len(peakL):
                ans.append(peakL[k])
                k+=1
        print(*ans)