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
lines = sys.stdin.readlines()
T = int(lines[0].strip())
for t in range(T):
    (n, k) = map(int, lines[2*t+1].strip().split(" "))
    nums = list(map(int, lines[2*t+2].strip().split(" ")))
    noChange = {}
    oneChange = {}
    for i in range(2, 2*k+4):
        noChange[i] = 0
        oneChange[i] = 0
    for i in range(n//2):
        l = nums[i]
        r = nums[-i-1]
        noChange[l+r] += 1
        oneChange[min(l,r)+1] += 1
        oneChange[l+r] -= 1
        oneChange[l+r+1] += 1
        oneChange[max(l,r)+k + 1] -= 1
    for i in range(3, 2*k+4):
        oneChange[i] += oneChange[i-1]
    
    res = n
    for i in range(2,2*k+3):
        tmp = n - oneChange[i] - 2*noChange[i]
        res = min(tmp, res)
    print(res)