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
from sys import stdin, stdout

K = 400006
freq = [0 for i in xrange(K)]

tc = input()
inp = stdin.readlines()
out = []

ptr = 0
while ptr < len(inp):
    n, k = map(int, inp[ptr].strip().split()); ptr += 1
    a = map(int, inp[ptr].strip().split()); ptr += 1

    # clear the freq array
    for i in xrange(k + k + 2): freq[i] = 0

    for i in xrange(n):
        j = n - 1 - i
        if j < i: break

        lo = 1 + min(a[i], a[j])
        hi = k + max(a[i], a[j])

        # no changes needed to achieve x = a[i] + a[j]
        # so, +1 in [lo, x) U [x + 1, hi]
        x = a[i] + a[j]

        # [lo, x)
        freq[lo] += 1
        freq[x]  -= 1

        # [x + 1, hi]
        freq[x + 1]  += 1
        freq[hi + 1] -= 1

        # trailing left and right parts

        # [2, lo)
        if lo > 2:
            freq[2] += 2
            freq[lo] -= 2

        # [hi + 1, k + k]
        if hi < k + k:
            freq[hi + 1] += 2
            freq[k + k + 1] -= 2

    ans = freq[2]
    for i in xrange(3, k + k + 1):
        freq[i] += freq[i - 1]
        ans = min(ans, freq[i])

    out.append(str(ans))

stdout.write("\n".join(out))