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:
'''
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

'''

for _ in range(int(input())):
    n, k = [int(i) for i in input().split()]
    a = [int(i) for i in input().split()]
    pairs = []
    psum = []
    pre1 = [0 for i in range(2 * k + 2)]
    pre0 = [0 for i in range(2 * k + 2)]
    for i in range(n // 2):
        pairs.append([a[i], a[n - i - 1]])
    for p in pairs:
        psum.append([min(p[0], p[1]) + 1, max(p[0], p[1]) + k])
        pre0[p[0] + p[1]] = pre0[p[0] + p[1]] + 1

    for mi, ma in psum:
        pre1[mi] = pre1[mi] + 1
        pre1[ma + 1] = pre1[ma + 1] - 1
    presum = 0

    for i in range(2, 2 * k + 2):
        presum = pre1[i] + presum
        pre1[i] = presum
    minch = n

    for i in range(2, 2 * k + 2):
        count0 = pre0[i]
        count1 = pre1[i] - count0
        count2 = n // 2 - count0 - count1

        total = 1 * count1 + 2 * count2
        minch = min(minch, total)
    #     print(minch)
    #
    # print(psum)
    # print(pre1)
    # print(pre0)
    print(minch)