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:
class BIT():
    def __init__(self, n):
        self.n = n
        self.bit = [0] * (n + 1)

    def _add(self, i, val):
        while i > 0:
            self.bit[i] += val
            i -= i & -i

    def get_val(self, i):
        i = i + 1
        s = 0
        while i <= self.n:
            s += self.bit[i]
            i += i & -i
        return s

    def add(self, l, r, val):
        self._add(r, val)
        self._add(l, -val)

import sys
input = sys.stdin.readline


t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    b = a[n // 2:n][::-1]
    a = a[0:n // 2]
    
    bit = BIT(2 * k + 1)
    bit.add(0, 2 * k + 1, (2 * n // 2))
    for i in range(n // 2):
        ran = [min(a[i] + 1, b[i] + 1), max(a[i] + k + 1, b[i] + k + 1)]
        bit.add(ran[0], ran[1], -1)
        bit.add(a[i] + b[i], a[i] + b[i] + 1, -1)
    print(min([bit.get_val(i) for i in range(2 * k + 1)]))