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:
for i in range(int(input())):
	n,k=map(int,input().split())
	map1={}
	pref=[]
	l=list(map(int,input().split()))
	for j in range(0,2*k+2):
		pref.append(0)
	for j in range(n//2):
		if(map1.get(l[j]+l[n-j-1])!=None):
			map1[l[j]+l[n-j-1]]+=1
		else:
			map1[l[j]+l[n-j-1]]=1
	summ=[]
	for j in range(n//2):
		summ.append((l[j],l[n-j-1]))
	for j in range(n//2):
		min_sum=min(summ[j][1],summ[j][0])+1
		max_sum=max(summ[j][1],summ[j][0])+k
		pref[min_sum]+=1
		pref[max_sum+1]-=1
	
	for j in range(3,2*k+2):
		pref[j]=pref[j-1]+pref[j]
	min_value=10000000
	
	for j in range(2,2*k+1):
		if(map1.get(j)!=None):
			a=map1[j]
		else:
			a=0
		b=pref[j]
		if(min_value>1*(b-a)+2*((n//2)-b)):
			min_value=1*(b-a)+2*((n//2)-b)
	print(min_value)