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:
from sys import setrecursionlimit
from collections import deque
import heapq
import functools
setrecursionlimit(10**6)
def IntInput(): return int(input())
def M_IntInput(): return map(int,input().split())
def M_StrInput(): return map(int,input().split())
def IntArrayInput(): return list(map(int,input().split()))
def modInverse(p,a,m): return ((p%m)*pow(a%m,m-2,m))%m
def compare(item1, item2): return item2[0] - item1[0] if item1[0]!=item2[0] else item1[1] - item2[1]
def DFS(s,n,paths,answer):
    visited = [False] * (n+1)
    stack = []
    stack.append(s) 
    visited[s] = True
    DFSArray = []
    while stack: 
        s = stack.pop()
        DFSArray.append(s)
        flag = 0
        for i in paths[s]: 
            if visited[i] == False: 
                flag=1
                stack.append(i) 
                visited[i] = True
    
        if not flag:
            answer[s-1][1]=0
    
    while DFSArray!=[]:
        s = DFSArray.pop()
        if answer[s-1][1]<0:
            answer[s-1][1] = 0
            for i in paths[s]:
                if answer[i-1][1]>=0:
                    answer[s-1][1] += answer[i-1][1]+1

    return answer
def BFS(s,n,paths,answer): 
    visited = [False] * (n+1)
    answer[s-1][0] = 0
    queue = deque()
    queue.append(s) 
    visited[s] = True
    while queue: 
        s = deque.popleft(queue)
        for i in paths[s]: 
            if visited[i] == False: 
                queue.append(i) 
                answer[i-1][0] = answer[s-1][0] + 1
                visited[i] = True
    return answer
numbers = ['0','1','2','3','4','5','6','7','8','9']
# ----------------------- Anshul Raj ------------------------ #

for _ in range(int(input())):
    n,k = M_IntInput()
    array = IntArrayInput()
    
    x = 0
    dictv = {}

    Nochange = [0]*(2*k + 1)
    OneChange = [0]*(2*k + 1)
    
    for i in range(n//2):
        Nochange[array[i] + array[n-i-1]]+=1
        OneChange[min(array[i],array[n-i-1])+1]+=1
        if max(array[i],array[n-i-1])!=k: OneChange[k + max(array[i],array[n-i-1]) + 1] -= 1


    sumV = 0
    ans = 100000000000
    for i in range(1,2*k+1):
        OneChange[i]+=OneChange[i-1]
        ans = min(ans,OneChange[i]-Nochange[i] + (n//2 - OneChange[i])*2)

    print(ans)