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:
for _ in range(int(input())):
    n,k=map(int,input().split())
    a=list(map(int,input().split()))
    l=[0]*(2*k+2)
    s=[0]*(2*k+2)
    coun=[]
    for i in range(n//2):
        s[a[i]+a[n-i-1]]+=1
        b=a[i]
        c=a[n-i-1]
        mi=min(a[i],a[n-i-1])
        ma=max(a[i],a[n-i-1])
        l[mi+1]+=1
        l[ma+k+1]-=1
    for i in range(2,len(l)):
        l[i]+=l[i-1]
    for i in range(2,2*k+1):
        count=l[i]-s[i]
        coun.append(count+2*(n//2-l[i]))
    print(min(coun))