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())
def oco(l):
    d = {}
    for el in l:
        if el in d:
            d[el] += 1
        else:
            d[el] = 1
    return d
for test in range(t):
    n, k = [int(i) for i in input().split()]
    a = [int(i) for i in input().split()]
    psum = [a[i] + a[n-i-1] for i in range(len(a)//2)]
    pmax = [k+max(a[i], a[n-i-1]) for i in range(len(a)//2)]
    pmax.sort()
    pmin = [1+min(a[i], a[n-i-1]) for i in range(len(a)//2)]
    pmin.sort()
    di = oco(psum)
    itma = 0
    itmi = 0
    ans = [0 for i in range(2*k-1)]
    for i in range(len(ans)):
        while itma < len(pmax):
            if pmax[itma]>=i+2:
                break
            itma += 1
        while itmi < len(pmin):
            if pmin[itmi]>i+2:
                break
            itmi += 1
        ans[i] = n +itma- itmi
        if i+2 in di:
            ans[i]  -= di[i+2]
    print(min(ans))