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:
# cook your dish here
from sys import stdin
import sys

if __name__=="__main__":
    for _ in range (int(stdin.readline())):
        n,k=map(int,(stdin.readline().split()))
        arr=list(map(int,stdin.readline().split()))
        count=[0]*((2*k)+1)
        for i in range (n//2):
            count[arr[i]+arr[n-1-i]]+=1
        prev=[0]*((2*k)+2)
        for i in range (n//2):
            left=min(arr[i],arr[n-1-i])+1
            right=max(arr[i],arr[n-i-1])+k
            prev[left]+=1
            prev[right+1]-=1
        for i in range (1,(2*k)+2):
            prev[i]+=prev[i-1]
        ans=1000000   
        for i in range (2,(2*k)+1):
            ans=min(ans,(prev[i]-count[i])+(n//2-prev[i])*2)   
        print(ans)