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 sys, math
input = sys.stdin.readline

def getInts():
    return [int(s) for s in input().split()]

def getInt():
    return int(input())

def getStrs():
    return [s for s in input().split()]

def getStr():
    return input().strip()

def listStr():
    return list(input().strip())

import collections as col
import math,itertools

"""
Every front and back pair must sum to the same
Suppose we have k = 6
1 1 1 3 5 6 2 4 2 2 Sums 3, 7, 11
Can we do it in 0 changes? No
1 change? Still no
2 changes? Maybe. 

6 1 1 7 6 3 4 6
aim for k+1
2,6 4,4 5,3 7,1
Worst case is n//2 - (pairs summing to k+1) because each pair can be mapped into k+1 in at most 1 step
Suppose we have a pair (sum p) that occurs q times. Then we have to change at least n//2-q digits. We then have to change an additional digit for
each pair where its minimum element > p-1, or where the max element is < p-K
We need to find the minimum such pair efficiently
Store a count of pairs by minimum element in a dictionary
Then create an array where the jth element is the number of pairs s.t. min(pair) >= j
Then create a dictionary of pair sums, and for each pair sum use the array to calculate the min steps
This should be O(K+N)
"""

def solve():
    N, K = getInts()
    A = getInts()
    pairs = col.defaultdict(int)
    min_el = col.defaultdict(int)
    max_el = col.defaultdict(int)
    for j in range(N//2):
        pairs[A[j]+A[N-1-j]] += 1
        min_el[min(A[j],A[N-1-j])] += 1
        max_el[max(A[j],A[N-1-j])] += 1
    #print(min_el)
    min_greater = [0]
    max_less = [0]
    for j in range(K,0,-1):
        min_greater.append(min_greater[-1]+min_el[j])
    min_greater.reverse()
    min_greater = [min_greater.pop()]+min_greater
    for j in range(1,K+1):
        max_less.append(max_less[-1]+max_el[j])
    #print(min_greater)
    #print(max_less)
    best = 10**9
    for pair in pairs:
        baseline = N//2-pairs[pair]
        #print(pair,baseline)
        #now add 1 for all the other pairs where the max element is P-K
        if pair-K > 1:
            baseline += max_less[pair-K-1]
        #if the pair is K+1 or greater, we don't need to worry about the minimum exceeding it
        if pair-1 < K:
            baseline += min_greater[pair]
        #print(pair,baseline)
        best = min(baseline,best)
    return min(best,N//2)
    
for _ in range(getInt()):
    print(solve())