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:
import sys

t_cases = int(input())

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

    # all a_i does not exceed some integer k
    # NO LIMIT ON X
    # thus the maximum sum of a_i + a_(n - i + 1) must be k + k (+ 1 because of array indexing)
    cnt = [0]*(2 * k + 1)
    for i in range(n//2):
        cnt[a[i] + a[n-i-1]] += 1
    # print(cnt)

    # the value pref[x] tells the number of such pairs that we need to replace at most one ELEMENT in this pair to make it sum equals x.

    # we can replace one element to fix this pair (i.e. to reach X) implies that the range must be min(pair) + 1 (because elements in the pair can't be 0) to max(pair) + k (no bigger than k); hence the largest possible value is

    pref = [0] * (2 * k + 2)
    for i in range(n//2):
        l1 = 1 + a[i]
        r1 = k + a[i]
        l2 = 1 + a[n-i-1]
        r2 = k + a[n-i-1]


        pref[min(l1, l2)] += 1
        pref[max(r1, r2) + 1] -= 1
        # print(str(i) + " " + str(pref))

    # print(pref)

    for i in range(1, len(pref) - 1):
        pref[i] += pref[i-1]

    res = sys.maxsize
    # minimum value is 2 because 1 + 1
    for i in range(2, 2*k+1):
        res = min(res, (pref[i] - cnt[i]) + (n//2 - pref[i])*2)
    print(res)