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:
import bisect
t = int(input())
for i in range(t):
    n, k = map(int, input().split())
    a = [int(i) for i in input().split()]
    ans = n // 2
    maxes = [0] * ans
    mines = [0] * ans
    summes = []
    count_summes = {}
    for i in range(n // 2):
        count_summes[a[i] + a[n - i - 1]] = 0
    for i in range(n // 2):
        maxes[i] = max(a[i], a[n - i - 1])
        mines[i] = min(a[i], a[n - i - 1])
        summes.append(a[i] + a[n - i - 1])
        count_summes[a[i] + a[n - i - 1]] += 1
    maxes.sort()
    mines.sort()
    for i in range(len(summes)):
        ans_now = 0
        s_now = summes[i]
        max_s = s_now - k
        min_s = s_now - 1
        first = 2 * (n // 2 - bisect.bisect_right(mines, min_s))
        second = 2 * bisect.bisect_left(maxes, max_s)
        ost = n // 2 - (first // 2) - (second // 2) - count_summes[s_now]
        ans = min(ans, ost + first + second)
    print(ans)