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:
a = int(input())
d = [0] * a
for i in range(0, a):
    b = int(input())
    d[i] = b
for i in range(0, a):
    m = 0
    l = 0
    f = [0] * d[i]
    if d[i] % 2 == 0:
        for i2 in range(1, d[i]//2+1):
            f[i2-1] = i2*2
            m += i2*2
            l += i2*2-1
            f[i2-1+(d[i]//2)] = i2*2-1
        if abs(m-l) % 2 == 0:
            if m > l:
                f[d[i]-1] = f[d[i]-1] + m-l
            else:
                f[d[i]//2-1] = f[d[i]//2-1] + l-m
            print("YES")
            print(*f)
        else:
            print("NO")
    else:
        print("NO")