Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> a(n);
    for (long long i = 0; i < n; ++i) {
      cin >> a[i];
    }
    long long size = 2 * k + 3;
    vector<long long> nul(size), one(size);
    for (long long i = 0; i < n / 2; ++i) {
      ++nul[a[i] + a[n - i - 1]];
      ++one[min(a[i], a[n - i - 1]) + 1];
      --one[a[i] + a[n - i - 1]];
      ++one[a[i] + a[n - i - 1] + 1];
      --one[max(a[i], a[n - i - 1]) + k + 1];
    }
    for (long long i = 1; i < size; ++i) {
      one[i] = one[i - 1] + one[i];
    }
    long long minn = n;
    for (long long i = 1; i <= 2 * k; ++i) {
      minn = min(minn, one[i] + (n / 2 - one[i] - nul[i]) * 2);
    }
    cout << minn << "\n";
  }
}