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 _ in range(int(input())):
	n, k = map(int,input().split())
	l = list(map(int,input().split()))
	cost = [0]*((2*k)+2)
	count  = [0]*((2*k)+2)
	for i in range(n//2):
		cost[min(l[i]+1,l[n-i-1]+1)]+=1
		cost[max(l[i]+k,l[n-i-1]+k)+1]-=1
		if min(l[i]+1,l[n-i-1]+1)-1 >= 2 :
			cost[2]+=2
			cost[min(l[i]+1,l[n-i-1]+1)]-=2
		if max(l[i]+k,l[n-i-1]+k)+1 <= 2*k:
			cost[max(l[i]+k,l[n-i-1]+k)+1]+=2
			cost[(2*k)+1]-=2
		cost[l[i]+l[n-i-1]]-=1
		cost[l[i]+l[n-i-1]+1]+=1
	for i in range(2,2*k+1):
		cost[i]=cost[i]+cost[i-1]
	ans=cost[(2*k)]
	for i in range(2,(2*k)+1):
		ans=min(ans,cost[i])
	print(ans)
# 2 1 2 2 3 2
# 2 2 2 1 2 1