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:
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)