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:
inp=input
li=list
iinp=lambda : int(inp())
ent=lambda : map(int,inp().split())
lient=lambda : [int(i) for i in inp().split()]
li0= lambda x: [0 for i in range(x)]
stent=lambda : [i for i in inp()]
mi=min
ma=max
l=len

t=iinp()
for _ in range(t):
    lon,valormax=ent()
    numeros=lient()
    s=[0]*(2*valormax+2)
    for i in range(lon//2):
        puntero1=numeros[i]
        puntero2=numeros[lon-i-1]
        s[2]+=2
        s[mi(puntero1,puntero2)+1]-=1
        s[ma(puntero1,puntero2)+valormax+1]+=1
        s[puntero1+puntero2]-=1
        s[puntero1+puntero2+1]+=1
    for i in range(2,(2*valormax+2)):
        s[i]+=s[i-1]
    print(mi(s[2:]))