Problem: You are given a positive integer n, it is guaranteed that n is even (i.e. divisible by 2).

You want to construct the array a of length n such that: 

  * The first n/2 elements of a are even (divisible by 2); 
  * the second n/2 elements of a are odd (not divisible by 2); 
  * all elements of a are distinct and positive; 
  * the sum of the first half equals to the sum of the second half (∑_{i=1}^{n/2} a_i = ∑_{i=n/2 + 1}^{n} a_i). 



If there are multiple answers, you can print any. It is not guaranteed that the answer exists.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of the array. It is guaranteed that that n is even (i.e. divisible by 2).

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — "NO" (without quotes), if there is no suitable answer for the given test case or "YES" in the first line and any suitable array a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) satisfying conditions from the problem statement on the second line.

Example

Input


5
2
4
6
8
10


Output


NO
YES
2 4 1 5
NO
YES
2 4 6 8 1 3 5 11
NO

Solution (in PYTHON3):

Solution:
from sys import stdin, exit
input = stdin.readline

def i(): return input()
def ii(): return int(input())
def iis(): return map(int, input().split())
def liis():	return list(map(int, input().split()))
def print_array(a): print(" ".join(map(str, a)))

t = ii()
for _ in range(t):
	n, k = iis()
	a = liis()
	counter = {}
	poss = []
	add = [0 for i in range(2*k+2)]
	rem = [0 for i in range(2*k+2)]
	alt = [0 for i in range(2*k+2)]
	
	for i in range(n//2):
		ini = a[i]	
		fim = a[n-i-1]
		mini = 1+min(ini, fim)
		maxi = k+max(ini, fim)
		alt[ini+fim] += 1
		add[mini] += 1
		rem[maxi+1] += 1
	
	count = 0
	for i in range(2, len(add)-1):
		count += add[i]-rem[i]
		two = ((n//2)-count)*2
		poss.append(count-alt[i]+two)
	
	print(min(poss))