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
#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 7;
long long res, n, m, cnt, ans, a, b, sum, mx, mn, c, d;
int main() {
  long long T;
  cin >> T;
  while (T--) {
    cin >> n >> m;
    vector<int> d(m * 2 + 2, 0);
    pair<long long, long long> v[n + 1];
    long long arr[n + 1];
    sum = ans = 0;
    for (int i = 1; i <= n; i++) {
      cin >> arr[i];
    }
    for (int i = 1; i <= n / 2; i++) {
      v[i] = {arr[i], arr[n - i + 1]};
    }
    d[1] = 2 * (n / 2);
    for (int i = 1; i <= n / 2; i++) {
      mn = min(1 + v[i].second, v[i].first + 1);
      mx = max(m + v[i].second, v[i].first + m);
      d[mn] -= 1, d[mx + 1] += 1;
      d[v[i].second + v[i].first] -= 1, d[v[i].second + v[i].first + 1] += 1;
    }
    sum = d[1], ans = sum;
    for (int i = 2; i <= m * 2; i++) {
      sum += d[i], ans = min(ans, sum);
    }
    cout << ans << endl;
  }
}
