Problem: You are given a positive integer n, it is guaranteed that n is even (i.e. divisible by 2).

You want to construct the array a of length n such that: 

  * The first n/2 elements of a are even (divisible by 2); 
  * the second n/2 elements of a are odd (not divisible by 2); 
  * all elements of a are distinct and positive; 
  * the sum of the first half equals to the sum of the second half (∑_{i=1}^{n/2} a_i = ∑_{i=n/2 + 1}^{n} a_i). 



If there are multiple answers, you can print any. It is not guaranteed that the answer exists.

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 only line of the test case contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of the array. It is guaranteed that that n is even (i.e. divisible by 2).

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

Output

For each test case, print the answer — "NO" (without quotes), if there is no suitable answer for the given test case or "YES" in the first line and any suitable array a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) satisfying conditions from the problem statement on the second line.

Example

Input


5
2
4
6
8
10


Output


NO
YES
2 4 1 5
NO
YES
2 4 6 8 1 3 5 11
NO

Solution (in PYTHON3):

Solution:
def r(s0, curr_min):
    count = 0
    for i in range(n // 2):
        s1 = a[i] + a[n - i - 1]

        if s1 > s0:
            t = s1 - s0
            if a[i] - t < 1 and a[n - i - 1] - t < 1:
                count += 2
            else:
                count += 1
        elif s1 < s0:
            t = s0 - s1
            if t + a[i] > k and t + a[n - i - 1] > k:
                count += 2
            else:
                count += 1
        if count > curr_min:
            break
    # print(s0, count)
    return count


for _ in range(int(input())):
    n, k = map(int, input().split())
    a = list(map(int, input().split()))

    cnt = [0] * (2 * k + 1)

    for i in range(n // 2):
        s = a[i] + a[n - i - 1]
        if s < 2 * k + 1:
            cnt[s] += 1

    milestones = [0] * (2 * k + 1 + 1)

    for i in range(n // 2):
        milestones[min(a[i], a[n - i - 1]) + 1] += 1
        milestones[max(a[i], a[n - i - 1]) + k + 1] += -1

    pref = [0] * (2 * k + 1)
    for i in range(1, 2 * k + 1):
        pref[i] = pref[i - 1] + milestones[i]

    count = n
    for i in range(2, 2 * k + 1):
        count = min(count, (pref[i] - cnt[i]) + (n // 2 - pref[i]) * 2)

    # print(cnt, pref)
    print(count)