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())
while t>0:
    x=int(input())
    nev=2+(x//2-1)*2
    sumev=((nev+2)//2)*((x//2))
    sumod=0
    k=1
    l2=[]
    for i in range(1,(x//2)+1):
        if i<(x//2):
            sumod+=k
            l2.append(k)
        if i==(x//2):
            if (sumev-sumod)%2!=0:



                l2.append(sumev-sumod)
                sumod+=(sumev-sumod)
            else:
                l2.append(0)
        k=k+2
    if 0 in l2:
        print('NO')
    if sumev==sumod:
        print('YES')
        z=2
        l1=[]
        for i in range(x//2):
            l1.append(z)
            z+=2
        print(*(l1+l2))
    t=t-1