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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
template <class T>
void clr(T& a, int n) {
  a.clear();
  a.resize(n + 1);
}
void solve();
const int N = 2e5 + 10;
const int INF = 1e9;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}
void rotate(vector<int>& a, int pos) {
  int temp = a[pos];
  a[pos] = a[pos + 2];
  a[pos + 2] = a[pos + 1];
  a[pos + 1] = temp;
}
int n;
void solve() {
  cin >> n;
  vector<int> a(n + 1);
  vector<pair<int, int> > b(n + 1);
  for (int i = int(1); i <= int(n); i++) {
    cin >> a[i];
    b[i] = {a[i], i};
  }
  vector<int> ans;
  sort((b).begin(), (b).end());
  for (int i = int(1); i <= int(n); i++) {
    a[b[i].second] = i;
  }
  int inv = 0;
  for (int i = int(1); i <= int(n); i++) {
    for (int j = int(1); j <= int(i - 1); j++) {
      inv += a[j] > a[i];
    }
  }
  bool uniq = 1;
  if (inv & 1) {
    for (int i = int(1); i <= int(n - 1); i++) {
      if (b[i].first == b[i + 1].first) {
        swap(a[b[i].second], a[b[i + 1].second]);
        uniq = 0;
        break;
      }
    }
    if (uniq) {
      cout << "-1\n";
      return;
    }
  }
  for (int i = int(1); i <= int(n - 2); i++) {
    int index = min_element(a.begin() + i, a.end()) - a.begin();
    while (index - i >= 2) {
      rotate(a, index - 2);
      ans.push_back(index - 2);
      index -= 2;
    }
    if (index - i) {
      rotate(a, i);
      rotate(a, i);
      ans.push_back(i);
      ans.push_back(i);
    }
  }
  cout << ans.size() << '\n';
  for (auto val : ans) cout << val << ' ';
  cout << '\n';
}