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
t = int(input())
for _ in range(t):
	n, k = list(map(int, input().split()))
	a = list(map(int, input().split()))
	sumFreq = []
	pref = []
	for i in range(2*k + 2):
		pref.append(0)
		sumFreq.append(0)
	i = 0
	j = n-1
	x = n//2
	while(x > 0):
		x -= 1
		s = a[i] + a[j]
		sumFreq[s] += 1
		p = k - min(a[i], a[j])
		q = max(a[i], a[j]) - 1
		r = [s - q, p + s]
		# print(r)
		pref[r[0]] += 1
		pref[r[1] + 1] -= 1
		i += 1
		j -= 1
	# print(sumFreq)
	for i in range(3, 2*k + 1):
		pref[i] += pref[i-1]
	# print(pref)
	ans = 1e9
	for i in range(2, 2*k + 1):
		ans = min(ans, (pref[i] - sumFreq[i]) + 2*(n//2 - pref[i]))
	print(ans)
	