Problem: You are given an array a consisting of n integers (it is guaranteed that n is even, i.e. divisible by 2). All a_i does not exceed some integer k.

Your task is to replace the minimum number of elements (replacement is the following operation: choose some index i from 1 to n and replace a_i with some integer in range [1; k]) to satisfy the following conditions:

  * after all replacements, all a_i are positive integers not greater than k; 
  * for all i from 1 to n/2 the following equation is true: a_i + a_{n - i + 1} = x, where x should be the same for all n/2 pairs of elements. 



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 first line of the test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k ≤ 2 ⋅ 10^5) — the length of a and the maximum possible value of some a_i correspondingly. It is guratanteed that n is even (i.e. divisible by 2). The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the i-th element of a.

It is guaranteed that the sum of n (as well as the sum of k) over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5, ∑ k ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the minimum number of elements you have to replace in a to satisfy the conditions from the problem statement.

Example

Input


4
4 2
1 2 1 2
4 3
1 2 2 1
8 7
6 1 1 7 6 3 4 6
6 6
5 2 6 1 3 4


Output


0
1
4
2

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))