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 PYTHON):

Solution:
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat
import math
import random

# sys.stdin = open('input')

def inp():
    return map(int, raw_input().split())

def inst():
    return raw_input().strip()


def my_main():
        t = inp()[0]
        for _ in range(t):
            n, k = inp()
            da = inp()
            mp = defaultdict(int)
            dmin = []
            dmax = []
            for i in range(n/2):
                a, b = da[i], da[n-i-1]
                mp[a+b] += 1
                dmin.append(min(a, b)+1)
                dmax.append(max(a, b)+k)
            dmin.sort()
            dmax.sort()
            ans = n/2
            n/=2
            for key, v in mp.iteritems():
                left = bisect.bisect_left(dmax, key)
                right = n - bisect.bisect_right(dmin, key)
                ans = min(ans, (left+right)*2+n-left-right-v)
            print ans









my_main()