You are given an array a consisting of n integers (it is guaranteed that n is even, i.e. divisible by 2). All a_i does not exceed some integer k.

Your task is to replace the minimum number of elements (replacement is the following operation: choose some index i from 1 to n and replace a_i with some integer in range [1; k]) to satisfy the following conditions:

  * after all replacements, all a_i are positive integers not greater than k; 
  * for all i from 1 to n/2 the following equation is true: a_i + a_{n - i + 1} = x, where x should be the same for all n/2 pairs of elements. 



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 first line of the test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k ≤ 2 ⋅ 10^5) — the length of a and the maximum possible value of some a_i correspondingly. It is guratanteed that n is even (i.e. divisible by 2). The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the i-th element of a.

It is guaranteed that the sum of n (as well as the sum of k) over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5, ∑ k ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the minimum number of elements you have to replace in a to satisfy the conditions from the problem statement.

Example

Input


4
4 2
1 2 1 2
4 3
1 2 2 1
8 7
6 1 1 7 6 3 4 6
6 6
5 2 6 1 3 4


Output


0
1
4
2
import sys
def solve():
    n,k=[int(x) for x in input().split()]
    li=[int(x) for x in input().split()]
    # how many pairs need 0 changes to have a particular sum
    zero_changes=[0]*(2*k+1)
    for i in range(n//2):
        zero_changes[li[i]+li[n-i-1]]+=1
    # how many pairs need 1 cheanges to have a particular sum
    one_changes=[0]*(2*k+2)    
    for i in range(n//2):
        maxp=max(li[i],li[n-i-1])+k
        minp=min(li[i],li[n-i-1])+1
        one_changes[minp]+=1
        one_changes[maxp+1]-=1
    # convert this optimised version of one_changes to what we actually need see algo notebook    
    for i in range(3,2*k+1):
        one_changes[i]+=one_changes[i-1]    
    # we will now calculate how many pair need 2 changes to have a particular sum along with the minimum number of changes 
    # so that all pairs have equal sum
    minm=sys.maxsize
    for i in range(2,2*k+1):
        zero=zero_changes[i]
        one=one_changes[i]-zero
        two=n//2-one-zero
        total_cost=2*two+1*one+0*zero
        minm=min(minm,total_cost)
    print(minm)    
    
    
t=int(input())
for i in range(t):
    solve()