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:
from sys import stdin, gettrace
from collections import Counter

if not gettrace():
    def input():
        return next(stdin)[:-1]


# def input():
#    return stdin.buffer.readline()

def main():
    def solve():

        n, k = map(int, input().split())
        aa = [int(a) for a in input().split()]
        saa = [0] * (n//2)
        mnv = [0] * (n//2)
        mxv = [0] * (n//2)
        for i in range(n//2):
            saa[i] = aa[i] + aa[-i-1]
            mnv[i] = min(aa[i], aa[-i-1]) + 1
            mxv[i] = max(aa[i], aa[-i-1]) + k
        saac = Counter(saa).most_common()
        saac.sort()
        mnv.sort()
        mxv.sort()
        mnp = 0
        mxp = 0
        best = n//2
        for s,c in saac:
            while mnp < n//2 and mnv[mnp] <= s:
                mnp += 1
            while mxp < n//2 and mxv[mxp] < s:
                mxp += 1
            opt = n - c - mnp + mxp
            best = min(best, opt)
        print(best)



    q = int(input())
    for _ in range(q):
        solve()


if __name__ == "__main__":
    main()