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:
import bisect
t = int(input())
for i in range(t):
    n, k = map(int, input().split())
    a = [int(i) for i in input().split()]
    ans = n // 2
    maxes = [0] * ans
    mines = [0] * ans
    summes = []
    count_summes = {}
    for i in range(n // 2):
        count_summes[a[i] + a[n - i - 1]] = 0
    for i in range(n // 2):
        maxes[i] = max(a[i], a[n - i - 1])
        mines[i] = min(a[i], a[n - i - 1])
        summes.append(a[i] + a[n - i - 1])
        count_summes[a[i] + a[n - i - 1]] += 1
    maxes.sort()
    mines.sort()
    for i in range(len(summes)):
        ans_now = 0
        s_now = summes[i]
        max_s = s_now - k
        min_s = s_now - 1
        first = 2 * (n // 2 - bisect.bisect_right(mines, min_s))
        second = 2 * bisect.bisect_left(maxes, max_s)
        ost = n // 2 - (first // 2) - (second // 2) - count_summes[s_now]
        ans = min(ans, ost + first + second)
    print(ans)