You are given an array a consisting of n integers (it is guaranteed that n is even, i.e. divisible by 2). All a_i does not exceed some integer k.

Your task is to replace the minimum number of elements (replacement is the following operation: choose some index i from 1 to n and replace a_i with some integer in range [1; k]) to satisfy the following conditions:

  * after all replacements, all a_i are positive integers not greater than k; 
  * for all i from 1 to n/2 the following equation is true: a_i + a_{n - i + 1} = x, where x should be the same for all n/2 pairs of elements. 



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 first line of the test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k ≤ 2 ⋅ 10^5) — the length of a and the maximum possible value of some a_i correspondingly. It is guratanteed that n is even (i.e. divisible by 2). The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the i-th element of a.

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

Output

For each test case, print the answer — the minimum number of elements you have to replace in a to satisfy the conditions from the problem statement.

Example

Input


4
4 2
1 2 1 2
4 3
1 2 2 1
8 7
6 1 1 7 6 3 4 6
6 6
5 2 6 1 3 4


Output


0
1
4
2
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)
	


