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;
class TaskF {
  int n;
  vector<int> res;
  vector<pair<int, int>> v;
  void rotate(int i) {
    res.emplace_back(i + 1);
    swap(v[i], v[i + 1]);
    swap(v[i], v[i + 2]);
  }

 public:
  TaskF(int n) {
    this->n = n;
    for (int x, i = 0; i < n; i++) {
      cin >> x;
      v.emplace_back(x, i);
    }
  }
  void solve() {
    int invs = 0;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        invs += (v[i] > v[j]);
      }
    }
    if (invs & 1) {
      auto impossible = [&]() {
        for (int i = 0; i < n; i++) {
          for (int j = i + 1; j < n; j++) {
            if (v[i].first == v[j].first) {
              swap(v[i], v[j]);
              return false;
            }
          }
        }
        return true;
      };
      if (impossible()) {
        cout << -1 << '\n';
        return;
      }
    }
    for (int i = 0; i < n - 2; i++) {
      pair<pair<int, int>, int> cur = {v[i], i};
      for (int j = i; j < n; j++) {
        cur = min(cur, {v[j], j});
      }
      while (cur.second > (i + 1)) {
        cur.second -= 2;
        rotate(cur.second);
      }
      if (cur.second != i) {
        rotate(i);
        rotate(i);
      }
    }
    cout << res.size() << '\n';
    for (auto &x : res) cout << x << ' ';
    cout << '\n';
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    TaskF F(n);
    F.solve();
  }
}