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:
import sys
# input = sys.stdin.buffer.readline
from collections import Counter
def I(): return(list(map(int,input().split())))
def sieve(n):
	a=[1]*n
	for i in range(2,n):
	    if a[i]:
	        for j in range(i*i,n,i):
	            a[j]=0
	return a
def calc(x):
	c=0
	n=len(arr)
	# print(arr)
	for i in range(n//2):
		if sumarr[i]!=x:
			m=max(arr[i],arr[n-i-1])
			if sumarr[i]<x:
				m=min(arr[i],arr[n-i-1])
				m+=x-sumarr[i]
				if m>k:c+=2
				else:c+=1
			else:
				m=max(arr[i],arr[n-i-1])
				m-=sumarr[i]-x
				if m<0:c+=2
				else:c+=1
	return c

for __ in range(int(input())):
	n,k=I()
	arr=I()
	sumarr=[0]*(n//2)
	for i in range(n//2):
		sumarr[i]=arr[i]+arr[n-i-1]
	# print(sumarr)
	c=Counter(sumarr)
	prefarr=[0]*(2*k+2)
	for i in range(n//2):
		a,b=max(arr[i],arr[n-i-1]),min(arr[i],arr[n-i-1])
		r=a+k
		l=b+1
		prefarr[l]+=1
		prefarr[r+1]-=1
		prefarr[a+b]-=1
		prefarr[a+b+1]+=1
	minans=n

	for i in range(2,2*k+1):
		prefarr[i]+=prefarr[i-1]
		minans=min(minans,prefarr[i]+2*(n//2-c.get(i,0)-prefarr[i]))
	# print(c)
	# print(prefarr)
	print(minans)