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:
for i in range(int(input())):
    n=int(input())
    if(n%4!=0):
        print('NO')
    else:    
        print('YES')
        m=n//2
        sumeven=m*(m+1)
        i=1
        sumodd=1
        l=[1]
        l1=[]
        for k in range(2,n+1,2):
            l1.append(k)
        #print(l1)
        while(sumeven!=sumodd):
            i=i+2 # 1 3
            sumodd=sumodd+i # 1 4 9 16 25 36 
            if(sumodd<sumeven):
                l.append(i)
                # 1 3 5 7
            elif(sumodd>sumeven):
                l.pop()
                sumodd -= (i-2)
                #print(sumodd,i)
                # 1 3 5 9
                l.append(i) # 1 3 5 11
            else:
                #l.pop() # 1 3 5 11
                l.append(i)
                break
            
        l3=l1+l
        #print(l)
        print(*l3,sep=' ')