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:
import sys
def solve():
    n,k=[int(x) for x in input().split()]
    li=[int(x) for x in input().split()]
    # how many pairs need 0 changes to have a particular sum
    zero_changes=[0]*(2*k+1)
    for i in range(n//2):
        zero_changes[li[i]+li[n-i-1]]+=1
    # how many pairs need 1 cheanges to have a particular sum
    one_changes=[0]*(2*k+2)    
    for i in range(n//2):
        maxp=max(li[i],li[n-i-1])+k
        minp=min(li[i],li[n-i-1])+1
        one_changes[minp]+=1
        one_changes[maxp+1]-=1
    # convert this optimised version of one_changes to what we actually need see algo notebook    
    for i in range(3,2*k+1):
        one_changes[i]+=one_changes[i-1]    
    # we will now calculate how many pair need 2 changes to have a particular sum along with the minimum number of changes 
    # so that all pairs have equal sum
    minm=sys.maxsize
    for i in range(2,2*k+1):
        zero=zero_changes[i]
        one=one_changes[i]-zero
        two=n//2-one-zero
        total_cost=2*two+1*one+0*zero
        minm=min(minm,total_cost)
    print(minm)    
    
    
t=int(input())
for i in range(t):
    solve()