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:
t = int(input())
for a in range(t):
    n,k = map(int,input().split())
    nums = list(map(int,input().split()))
    lookUp = dict()
    sumMinMax = []
    values  = [0 for c in range(2*k+2)]
    for b in range(n//2):
        pairSum = nums[b] + nums[n-b-1]
        if pairSum in lookUp:
            lookUp[pairSum] += 1
        else:
            lookUp[pairSum] = 1
        sumMinMax.append([pairSum, 1 + min(nums[b],nums[n-b-1]), k + max(nums[b],nums[n-b-1])])
        values[1 + min(nums[b],nums[n-b-1])] += 1
        values[k + max(nums[b],nums[n-b-1])+1] -= 1
    sumPrevious = 0
    for c in range(len(values)):
        values[c] += sumPrevious
        sumPrevious = values[c]
    best = 0
    for key in lookUp:
        actual = lookUp[key] - (n//2 - values[key])
        if actual > best:
            best = actual
    print(n//2-best)