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
import sys
range = xrange
input = raw_input

inp = [int(x) for x in sys.stdin.read().split()]; ii = 0
t = inp[ii]; ii += 1
out = []
for _ in range(t):
    n = inp[ii]; ii += 1
    k = inp[ii]; ii += 1
    A = inp[ii: ii + n]; ii += n

    for i in range(n//2):
        A[i], A[~i] = min(A[i],A[~i]), max(A[i], A[~i])

    too_small = [0]*(2*k + 2)
    too_large = [0]*(2*k + 2)
    equal = [0]*(2*k + 1)
    for i in range(n//2):
        a = A[i]
        b = A[~i]
        equal[a + b] += 1
        too_small[a] += 1
        too_large[b + k + 1] += 1

    x = 0
    for j in range(len(too_large)):
        x += too_large[j]
        too_large[j] = x

    x = 0
    for j in reversed(range(len(too_small))):
        x += too_small[j]
        too_small[j] = x

    best = 2 * n
    for x in range(2, 2 * k + 1):
        # 0: equal
        # 1: n//2 - too_large - too_small - equal
        # 2: too_large + too_small
        best = min(best, n//2 + too_large[x] + too_small[x] - equal[x])
    print best
