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:
import io, os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

tokens = []
tokens_next = 0


def next_str():
    global tokens, tokens_next
    while tokens_next >= len(tokens):
        tokens = input().split()
        tokens_next = 0
    tokens_next += 1
    return tokens[tokens_next - 1]


def nextInt():
    return int(next_str())


# the idea here is that, each pair has a range where it needs to only change one of the two numbers in it
# to match a certain sum and outside of that range u need to change the 2 numbers to match that sum
# (except for pair[0] + pair[1] cz u don't need to change anything)
# so for each possible sum we try to calculate the number of elements to change to obtain that sum
# if we do that for each pair individually we get smth like this: 2 2 2 1 1 0 1 1 2 2 2
# we try to aggregate arrays like this over all pairs
# we first compress the above array to its diff: 2 0 0 -1 0 +1 0 0
# and to reduce time complexity we accumulate the changes for all pairs
# so we basically accumulate all the "diff" arrays
# and then we calculate the final  "changes" array but "decompressing" the diff

t = nextInt()
for _ in range(t):
    n = nextInt()
    k = nextInt()
    arr = [nextInt() for i in range(n)]

    changes = [0] * (2 * k + 10)
    for i in range(n // 2):
        a = arr[i]
        b = arr[n - i - 1]

        # inclusive
        st = min(a, b) + 1
        end = max(a, b) + k

        if st > 2:
            changes[2] += 2
            changes[st] -= 1
        else:
            changes[2] += 1

        if end + 1 < len(changes):
            changes[end + 1] += 1
        changes[a + b] -= 1
        changes[a + b + 1] += 1

    for i in range(3, len(changes)):
        changes[i] += changes[i - 1]

    print(min(changes[2:]))