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:
test_case = int(input())
array = []
even_sum = 0
odd_sum = -1
while test_case > 0:
    length = int(input())
    even_half = length // 2
    for x in range(even_half):
        even_sum = even_sum + 2
        array.append(even_sum)
    for y in range(even_half - 1):
        odd_sum = odd_sum + 2
        array.append(odd_sum)

    remain = sum(array[even_half:])

    if (sum(array[0: even_half]) - remain) % 2 != 0:
        array.append(sum(array[0: even_half]) - remain)
        print("YES")
        print(*array, sep=" ")
    else:
        print("NO")

    test_case -= 1
    even_sum = 0
    odd_sum = -1
    array = []