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:
def rangesum(n):
    s=0
    i=2
    c=0
    even=[]
    while c<n//2:
        s+=i
        even.append(i)
        i+=2
        c+=1
    i=1
    c=0
    odd=[]
    s2=0
    while c<n//2:
        if s2>s:
            break
            #i+=2
        odd.append(i)
        s2+=i
        i+=2
        c+=1
    if (s-sum(odd[0:n//2-1]))%2!=0:
        odd[-1]=s-sum(odd[0:n//2-1])
    if sum(odd)!=sum(even):
        print("NO")
    else:
        print("YES")
        print(*(even+odd))
    

for _ in range(int(input())):
    n=int(input())
    rangesum(n)