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:
import sys
range = xrange
input = raw_input

inp = [int(x) for x in sys.stdin.read().split()]; ii = 0
t = inp[ii]; ii += 1
out = []
for _ in range(t):
    n = inp[ii]; ii += 1
    k = inp[ii]; ii += 1
    A = inp[ii: ii + n]; ii += n

    for i in range(n//2):
        A[i], A[~i] = min(A[i],A[~i]), max(A[i], A[~i])

    too_small = [0]*(2*k + 2)
    too_large = [0]*(2*k + 2)
    equal = [0]*(2*k + 1)
    for i in range(n//2):
        a = A[i]
        b = A[~i]
        equal[a + b] += 1
        too_small[a] += 1
        too_large[b + k + 1] += 1

    x = 0
    for j in range(len(too_large)):
        x += too_large[j]
        too_large[j] = x

    x = 0
    for j in reversed(range(len(too_small))):
        x += too_small[j]
        too_small[j] = x

    best = 2 * n
    for x in range(2, 2 * k + 1):
        # 0: equal
        # 1: n//2 - too_large - too_small - equal
        # 2: too_large + too_small
        best = min(best, n//2 + too_large[x] + too_small[x] - equal[x])
    print best