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;
long long n = 0;
long long arr[501];
void reset() {}
void init() {}
pair<bool, vector<long long>> solve_vector(
    vector<pair<long long, long long>> v) {
  vector<long long> res;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < v.size(); j++) {
      if (v[j].second > j) {
        while (v[j].second - j >= 2) {
          res.push_back(v[j].second - 2);
          long long ind1 = 0, ind2 = 0;
          long long ind3 = j;
          for (long long k = 0; k < v.size(); k++) {
            if (v[k].second == v[j].second - 2) ind1 = k;
            if (v[k].second == v[j].second - 1) ind2 = k;
          }
          v[ind1].second++;
          v[ind2].second++;
          v[ind3].second -= 2;
        }
        if (v[j].second - j == 1) {
          res.push_back(v[j].second - 1);
          res.push_back(v[j].second - 1);
          long long ind1 = -1, ind2 = j;
          long long ind3 = -1;
          for (long long k = 0; k < v.size(); k++) {
            if (v[k].second == v[j].second - 1) ind1 = k;
            if (v[k].second == v[j].second + 1) ind3 = k;
          }
          if (ind3 == -1) return {false, vector<long long>()};
          v[ind1].second = j + 2;
          v[ind3].second = j + 1;
          v[j].second = j;
        }
        break;
      }
    }
  }
  return {true, res};
}
void solve() {
  vector<pair<long long, long long>> v1, v2;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    v1.push_back({arr[i], i});
    v2.push_back({arr[i], i});
  }
  sort(v1.begin(), v1.end());
  sort(v2.begin(), v2.end());
  bool flag = false;
  for (long long i = 0; i < v2.size(); i++) {
    for (long long j = 0; j < v2.size(); j++)
      if (i != j && v2[i].first == v2[j].first) {
        swap(v2[i], v2[j]);
        flag = true;
        break;
      }
    if (flag) break;
  }
  pair<bool, vector<long long>> p1 = solve_vector(v1);
  pair<bool, vector<long long>> p2 = solve_vector(v2);
  if (p1.first) {
    cout << p1.second.size() << endl;
    for (long long i = 0; i < p1.second.size(); i++)
      cout << p1.second[i] + 1 << ' ';
    cout << endl;
  } else if (p2.first) {
    cout << p2.second.size() << endl;
    for (long long i = 0; i < p2.second.size(); i++)
      cout << p2.second[i] + 1 << ' ';
    cout << endl;
  } else
    cout << -1 << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  init();
  long long t = 0;
  cin >> t;
  while (t--) {
    solve();
    reset();
  }
  return 0;
}