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 PYTHON):

Solution:
from sys import stdin, stdout

K = 400006
freq = [0 for i in xrange(K)]

tc = input()
inp = stdin.readlines()
out = []

ptr = 0
while ptr < len(inp):
    n, k = map(int, inp[ptr].strip().split()); ptr += 1
    a = map(int, inp[ptr].strip().split()); ptr += 1

    # clear the freq array
    for i in xrange(k + k + 2): freq[i] = 0

    for i in xrange(n):
        j = n - 1 - i
        if j < i: break

        lo = 1 + min(a[i], a[j])
        hi = k + max(a[i], a[j])

        # no changes needed to achieve x = a[i] + a[j]
        # so, +1 in [lo, x) U [x + 1, hi]
        x = a[i] + a[j]

        # [lo, x)
        freq[lo] += 1
        freq[x]  -= 1

        # [x + 1, hi]
        freq[x + 1]  += 1
        freq[hi + 1] -= 1

        # trailing left and right parts

        # [2, lo)
        if lo > 2:
            freq[2] += 2
            freq[lo] -= 2

        # [hi + 1, k + k]
        if hi < k + k:
            freq[hi + 1] += 2
            freq[k + k + 1] -= 2

    ans = freq[2]
    for i in xrange(3, k + k + 1):
        freq[i] += freq[i - 1]
        ans = min(ans, freq[i])

    out.append(str(ans))

stdout.write("\n".join(out))