We define $\mbox{P}$ to be a permutation of the first $n$ natural numbers in the range $[1,n]$. Let $p o s[i]$ denote the value at position $\boldsymbol{i}$ in permutation $\mbox{P}$ using $1$-based indexing. 

$\mbox{P}$ is considered to be an absolute permutation if  $|pos[i]-i|=k$ holds true for every $i\in[1,n]$.

Given $n$ and $\boldsymbol{\mbox{k}}$, print the lexicographically smallest absolute permutation $\mbox{P}$.  If no absolute permutation exists, print -1.

Example 

$n=4$ 

$k=2$   

Create an array of elements from $1$ to $n$, $pos=[1,2,3,4]$.  Using $1$ based indexing, create a permutation where every $|pos[i]-i|=k$.  It can be rearranged to $[3,4,1,2]$ so that all of the absolute differences equal $k=2$:  

pos[i]  i   |pos[i] - i|
  3     1        2
  4     2        2
  1     3        2
  2     4        2

Function Description  

Complete the absolutePermutation function in the editor below.  

absolutePermutation has the following parameter(s):  

int n: the upper bound of natural numbers to consider, inclusive  
int k: the absolute difference between each element's value and its index  

Returns   

int[n]: the lexicographically smallest permutation, or $\left[{-1}\right]$ if there is none   

Input Format

The first line contains an integer $\boldsymbol{\boldsymbol{t}}$, the number of queries. 

Each of the next $\boldsymbol{\boldsymbol{t}}$ lines contains $2$ space-separated integers, $n$ and $\boldsymbol{\mbox{k}}$.  

Constraints

$1\leq t\leq10$
$1\leq n\leq10^5$
$0\leq k<n$

Sample Input
STDIN   Function
-----   --------
3       t = 3 (number of queries)
2 1     n = 2, k = 1
3 0     n = 3, k = 0
3 2     n = 3, k = 2

Sample Output
2 1
1 2 3
-1

Explanation

Test Case 0:

Test Case 1:

Test Case 2: 

No absolute permutation exists, so we print -1 on a new line.

Difficulty: MEDIUM

Solutions:
import sys
f = [i.strip() for i in sys.stdin.readlines()]
for t in range(1, len(f)):
	N = int(f[t].split(' ')[0])
	k = int(f[t].split(' ')[-1])
	A = []
	S = set()
	for pos in range(1, N + 1):
		if pos in S:
			a = pos + k
			if a <= N:
				A.append(a)
				S.remove(pos)
				pos2 = pos + 2 * k
				if pos2 <= N:
					S.add(pos2)
			else:
				break
		else:
			a = pos - k
			b = pos + k
			if a >= 1:
				A.append(a)
			elif b <= N:
				A.append(b)
				pos2 = pos + 2 * k
				if pos2 <= N:
					S.add(pos2)
			else:
				break
	else:
		print(' '.join([str(i) for i in A]))
		continue
	print('-1')