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:
def inp():
    return (int(input()))


def inlt():
    return (list(map(int, input().split())))


def insr():
    s = input()
    return (list(s[:len(s) - 1]))


def invr():
    return (map(int, input().split()))


def findmin(x, y):
    if x < y:
        return x
    return y


def findmax(x, y):
    if x > y:
        return x
    return y

t = inp()
for i in range(0, t):
    n, k = invr()
    a = list(invr())
    min, max = 0, 0
    f = [0]*(2*k+2)
    cnt = [0]*(2*k+1)
    for j in range(0, int(n / 2)):
        cnt[a[j] + a[n - j - 1]] = cnt[a[j] + a[n - j - 1]] + 1
        min = findmin(a[j], a[n - j - 1]) + 1
        max = findmax(a[j], a[n - j - 1]) + k
        f[min]=f[min]+1
        f[max+1]=f[max+1]-1

    for j in range(1, 2*k+1):
        f[j] = f[j]+f[j-1]

    result = 999999
    for l in range(2, 2 * k + 1):
        q = f[l] - cnt[l] + (n / 2 - f[l]) * 2
        if result > q:
            result = q
    print(int(result))