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:
t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    freq = [0 for i in range(2*k+5)]
    freq2 = [0 for i in range(2*k+5)]
    def check(ans1):
        ans = float('inf')
        best = 0
        for i in range(1, (2*k)+5):
            tmp = 0
            for j in range(n//2):
                if (a[j] + a[n-j-1] != i):
                    if (i-a[j] >= 1 and i - a[j] <= k):
                        tmp += 1
                    elif (i-a[n-j-1] >= 1 and i-a[n-j-1] <= k):
                        tmp += 1
                    else:
                        tmp += 2
            if (tmp < ans):
                best= i
            ans = min(ans, tmp)
        return ans
    for i in range(n//2):
        freq[min(a[n-1-i], a[i])+1] -= 1
        freq[0] += 1
        freq[max(a[n-1-i], a[i])+k+1] += 1
    for i in range(n//2):
        freq2[a[i] + a[n-i-1]] +=1
    ans = float('inf')
    for i in range(1, 2 * k + 1):
        freq[i] += freq[i-1]
        ans = min(ans, freq[i] + n//2 - freq2[i])
    print(ans)