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
input=sys.stdin.readline
from collections import defaultdict as dd,deque as dq
t=int(input())
import bisect
while t:
    d=dd(int)
    n,k=map(int,input().split())
    l=list(map(int,input().split()))
    ll=[]
    mx=1000000000000000
    mi=-1
    ll=[]
    l2=[]
    ri=-1
    le=mx
    for i in range(n//2):
        d[l[i]+l[n-i-1]]+=1
        
        mx=min(mx,max(l[i],l[n-i-1]))
        mi=max(mi,min(l[i],l[n-i-1]))
        ll.append(max(l[i],l[n-i-1])+k)
        l2.append(min(l[i],l[n-i-1])+1)
        le=min(le,l[i]+l[n-i-1])
        ri=max(ri,l[i]+l[n-i-1])
    res=n//2
    ll.sort()
    l2.sort()
    ri=ll[-1]
    if(le>ri):
        le,ri=ri,le
    #print(mi,mx,le,ri)
    mx=0
    ind=100000000000
    #print(ll,le,ri)
    #print(mx)
    #print(l2)
    for i in range(le,ri+1):
        cou=bisect.bisect_left(ll,i)
        cou2=n//2-bisect.bisect_right(l2,i)
        #print(cou,cou2,i)
        if(mx<d[i]-cou-cou2):
            #print("lol",d[i],cou,i)
            mx=d[i]-cou-cou2
            #ind=min(ind,i)
    cou=0
    print(res-mx)
    t-=1