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 main():
    t = int(input())
    while t > 0:
        n, k = map(int, input().split())
        arr = [int(e) for e in input().split()]
        mp = {}
        vec = []
        for i in range(int(n/2)):
            if arr[i] + arr[n-i-1] not in mp:
                mp[arr[i]+arr[n-i-1]] = 0
            mp[arr[i]+arr[n-1-i]] += 1
            vec.append([min(arr[i], arr[n-i-1]), max(arr[i], arr[n-i-1])])

        pre = [0 for i in range(2*k+10)]

        for f, s in vec:
            l = f + 1
            r = s + k
            pre[l] += 1
            pre[r+1] -= 1

        sum = 0
        for i in range(0, 2*k+10):
            sum += pre[i]
            pre[i] = sum

        ans = n
        for x in range(2, 2*k+1):
            cnt0 = mp.get(x)
            if cnt0 == None:
                cnt0=0;
            cnt1 = pre[x]-cnt0
            cnt2 = n/2-cnt0-cnt1
            ans = min(ans, cnt1+2*cnt2)

        print(int(ans))
        t -= 1

if __name__ == "__main__":
    main()