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:
for _ in range(int(input())):
    n, k = map(int, input().split())
    if(n < 3 and k == 0):
        print(' '.join(map(str, [i for i in range(1, n + 1)])))
    elif(n < 3):
        print(-1)
    else:
        if(n % 2 == 0 and k <= n // 2 - 1):
            ans = [i for i in range(n + 1)]
            count = 0
            if(k > 0):
                for i in range(2, n):
                    if(i % 2 == 0):
                        ans[i] += 1
                    else:
                        ans[i] -= 1
                    count += 1
                    if(count == k * 2):
                        break
            print(*ans[1:])
        elif(n % 2 == 0):
            print(-1)
        elif(n % 2 == 1 and k <= n // 2):
            ans = [i for i in range(n + 1)]
            count = 0
            if(k > 0):
                for i in range(2, n + 1):
                    if(i % 2 == 0):
                        ans[i] += 1
                    else:
                        ans[i] -= 1
                    count += 1
                    if(count == k * 2):
                        break
            print(*ans[1:])
        else:
            print(-1)