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 sys
import math

def ii():
    return sys.stdin.readline().strip()

def idata():
    return [int(x) for x in ii().split()]

def bfs(graph1, root1):
    visited1 = set()
    queue1 = [[root1, 0]]
    visited1.add(root1)
    qw, ver = 0, 1
    while queue1:
        vertex1 = queue1[0]
        queue1 = queue1[1:]
        for neig1 in graph1[vertex1[0]]:
            if neig1 not in visited1:
                visited1.add(neig1)
                queue1 += [[neig1, vertex1[1] + 1]]
                if qw < vertex1[1] + 1:
                    ver, qw = neig1, vertex1[1] + 1
    return ver, qw

def dfs(graph, ver):
    used_v = set()
    depth = [0] * (len(graph) + 1)
    level = [ver]
    num_of_level = 0
    while level:
        new_level = []
        for v in level:
            depth[v] = num_of_level
            used_v.add(v)
            for v1 in graph[v]:
                if v1 not in used_v:
                    new_level += [v1]
        level = new_level
        num_of_level += 1
    return depth

def binary_search(array, x):
    left, right = -1, len(array)
    while left + 1 != right:
        middle = (left + right) // 2
        if array[middle] >= x:
            right = middle
        elif array[middle] < x:
            left = middle
    return right

def c_mod(n1, k1, mod1):
    num = den = 1
    for i in range(n1 - k1):
        num = (num * (n1 - i)) % mod1
        den = (den * (i + 1)) % mod1
    return (num * pow(den, mod1 - 2, mod1)) % mod1

def c(n1, k1):
    num = den = 1
    for i in range(n1 - k1):
        num = num * (n1 - i)
        den = den * (i + 1)
    return num // den

def v_sistemu(x, k):
    x = int(x)
    z = ''
    while x:
        z += str(x % k)
        x //= k
    return z[::-1]

def iz_sistemi(x, k):
    x = str(x)[::-1]
    ans = 0
    for i in range(len(x)):
        ans += int(x[i]) * pow(k, i)
    return ans

def solve_of_problem():
    n = int(ii())
    if n % 4:
        print('NO')
    else:
        print('YES')
        a, b = [], []
        for i in range(n // 2):
            a += [(i + 1) * 2]
            b += [i * 2 + 1]
        b[-1] += n // 2
        print(*a, *b, sep=' ')
    return

for ______ in range(int(ii())):
    solve_of_problem()