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:
t = int(input())

for _ in range(t):
    [n, k] = map(int, input().split())
    a = [*map(int, input().split())]

    cnt = [0] * (2*k)
    intervals = [0] * (2*k)
    for i in range(n//2):
        v = a[i] + a[n-i-1]

        cnt[v-1] += 1

        ma = max(a[i], a[n-i-1])
        mi = min(a[i], a[n-i-1])

        intervals[mi] += 1
        if ma+k < 2*k:
            intervals[ma+k] -= 1 # -1 at index i+1 to "cancel" the value during prefix sum 

    # prefix sum => thaks to that we are not O(N*K) but O(N + K)
    for i in range(1, 2*k):
        intervals[i] += intervals[i-1]

    ans = 1e9
    for i in range(1, 2*k):
        # min(ans, modif en 1 - modif en 0 + modifs en 2 [=paires - modif en 1 * 2])
        ans = min(ans, intervals[i] - cnt[i] + (n//2 - intervals[i]) * 2)
    
    print(ans)