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:
#1343D
#Constant Palindrome Sum
#len(Array)%2==0
#Replace the minimum number of elements to satisfy:
#1. for all i from 1 to n/2 the following equation is 
#true : ai+a(n-i+1)==x, where x should be the same #for all n/2 pairs of elements

def D():
	T=int(input())
	while(T!=0):
		T=T-1
		n,k=map(int,input().split())
		Arr=[int(i) for i in input().split(" ")]
		FreqArr=[0]*((2*k)+1)
		d=len(Arr)//2
		maxctn=0
		for i in range(0,d):
			index=Arr[i]+Arr[n-1-i]
			FreqArr[index]+=1
			if(FreqArr[index]>maxctn):
				maxctn=FreqArr[index]
		
		if(maxctn==(n//2)):
			print(0)
			continue

		Pref=[0]*((2*k)+2)
		for j in range(0,d):
			MinSum=min(Arr[j],Arr[n-1-j])+1
			MaxSum=max(Arr[j],Arr[n-1-j])+k
			Pref[MinSum]+=1
			Pref[MaxSum+1]-=1
			#print(Pref)
		#print(Pref)
		ans=1e11
		for i in range(2,(2*k)+1):
			Pref[i]=Pref[i-1]+Pref[i]
			val1=Pref[i]-FreqArr[i]
			val2=2*(d-Pref[i])
			ans=min(ans,val1+val2)
		print(ans)

if __name__ == '__main__':
	D()