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:
# map(int, input().split())
t = int(input())
for _ in range(t):
    n, k = map(int, input().split())

    a = list(map(int, input().split()))

    lst = [0] * (2*k + 2)
    interval = []
    for i in range(n//2):
        st = min(a[i], a[n-1-i])
        s = a[i] + a[n-1-i]
        end = s - st + k
        lst[st + 1] +=1
        lst[end + 1] -= 1
        # add = 1 if st + 1 != s else 0
        lst[s] += 1
        lst[s+1] -= 1
    cur = 0
    idx = 0
    interval = [0, 0]
    maxy = -1
    for i in range(2*k+1):
        cur += lst[i] 
        if maxy<cur : 
            maxy=cur 
            idx=i  
            interval[0] = i
        if lst[i] < 0:
            interval[1] = i-1
    res = idx
    # res = lst.index(max(lst))
    # print(lst)
    # print(res)
    ans = 0
    for i in range(n//2):
        st = min(a[i], a[n-1-i])
        s = a[i] + a[n-1-i]
        end = s - st + k
        if res == s:
            continue
        elif st+1 <= res <= end:
            # st <= interval[0] and interval[1] <= end
            # print("This is i", i)
            ans+=1
        else:
            # print("This is 2i", i)
            ans+=2
    print(ans)


# def maxOverlap(start,end): 
   
#     n= len(start) 
#     maxa = max(start)# Finding maximum starting time 
#     maxb = max(end)  # Finding maximum ending time 
#     maxc=max(maxa,maxb) 
#     x =(maxc+2)*[0] 
#     cur=0; idx=0
   
#     for i in range(0,n) :# CREATING AN AUXILIARY ARRAY 
#         x[start[i]]+=1 # Lazy addition 
#         x[end[i]+1]-=1
        
#     maxy=-1
#     #Lazily Calculating value at index i 
#     for i in range(0,maxc+1):  
#         cur+=x[i] 
#         if maxy<cur : 
#             maxy=cur 
#             idx=i      
#     print("Maximum value is: {0:d}".format(maxy), 
#                      " at position: {0:d}".format(idx))