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
for t in range(int(input())):
	n, k = [int(j) for j in input().split()]
	l = [int(j) for j in input().split()]
	a = 0
	b = 0
	s = [0]*(2*k+1)
	t = [0]*(2*k+1)
	pre = [0]*(2*k)
	suf =  [0]*(2*k)
	for i in range(n//2):
		tmp = l[i]+l[n-i-1]
		s[tmp] += 1
		pre[min(l[i], l[n-i-1])]+=1
		suf[max(l[i], l[n-i-1])]+=1
	# pre = []*(2*k)
	# suf =  [s[n-1]]*(2*k)
	# print(suf)
	for i in range(1,len(pre)):
		pre[i] += pre[i-1]
		suf[i] += suf[i-1]
	# print(suf)
	# print(pre)
	# suf = suf[::-1]
	ans = n
	# print(s, suf, pre)
	for sum_ in range(1,2*k+1):
		tmp = suf[max(sum_-k-1, 0)]+n//2-s[sum_]+n//2-pre[max(sum_-1, 0)]
		# print(tmp, sum_, pre[max(sum_-2, 0)], suf[max(sum_-k-1, 0)], s[sum_])
		# print(tmp)
		ans = min(ans, tmp)
	print(ans)


