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:
# WeirdBugsButOkay

def solve() :
    n, k = list(map(int, input().split()))
    a = list(map(int, input().split()))
    cnt = [0] * (2 * k + 2)
    sum = [0] * (2 * k + 2)
    for i in range (0, n // 2) :
        sum[a[i] + a[n - 1 - i]] += 1
    for i in range (0, n // 2) :
        cnt[min(a[i], a[n - 1 - i]) + 1] += 1
        cnt[max(a[i], a[n - 1 - i]) + k + 1] -= 1
    for i in range (1, 2 * k + 1) :
        cnt[i] += cnt[i - 1]
    ans = n
    for i in range (2, 2 * k + 1) :
        sc = cnt[i]
        nc = sum[i]
        dc = n // 2 - cnt[i]
        ans = min(ans, sc + 2 * dc - nc)
    print(ans)

t = int(input())
for _ in range (0, t) :
    solve()