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;
vector<string> vec_splitter(string s) {
  for (char& c : s) c = c == ',' ? ' ' : c;
  stringstream ss;
  ss << s;
  vector<string> res;
  for (string z; ss >> z; res.push_back(z))
    ;
  return res;
}
void debug_out(vector<string> args, int idx) { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, Head H, Tail... T) {
  if (idx > 0) cerr << ", ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, T...);
}
void localTest() {}
void doit(vector<long long>& A, long long idx) {
  swap(A[idx + 1], A[idx + 2]);
  swap(A[idx], A[idx + 1]);
  return;
}
int main() {
  localTest();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> A(n), ans;
    vector<pair<long long, long long> > B(n);
    for (long long i = 0; i < n; ++i) {
      cin >> A[i];
      B[i] = {A[i], i};
    }
    sort((B).begin(), (B).end());
    for (long long i = 0; i < n; ++i) {
      A[B[i].second] = i;
    }
    long long inversions = 0;
    for (long long i = 0; i < n; ++i) {
      for (long long j = 0; j < i; ++j) {
        if (A[j] > A[i]) ++inversions;
      }
    }
    if (inversions & 1) {
      for (long long i = 0; i < n - 1; ++i) {
        if (B[i].first == B[i + 1].first) {
          swap(A[B[i].second], A[B[i + 1].second]);
          break;
        }
      }
    }
    for (long long i = 0; i < n - 2; ++i) {
      long long pos = min_element(A.begin() + i, A.end()) - A.begin();
      while (pos > i + 1) {
        doit(A, pos - 2);
        ans.push_back(pos - 2);
        pos -= 2;
      }
      if (pos != i) {
        doit(A, i);
        doit(A, i);
        ans.push_back(i);
        ans.push_back(i);
      }
    }
    if (!is_sorted((A).begin(), (A).end()))
      cout << -1 << "\n";
    else {
      cout << ans.size() << "\n";
      for (auto u : ans) cout << u + 1 << " ";
      cout << "\n";
    }
  }
  return 0;
}