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;
const long long MAX = 1e6 + 8;
const long long MOD = 1e9 + 7;
const long long sz = 1e5 + 5;
void f_io() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
}
int32_t main() {
  f_io();
  long long tc = 1;
  cin >> tc;
  auto rotate = [](vector<long long>& v, long long ind) {
    swap(v[ind + 1], v[ind + 2]);
    swap(v[ind], v[ind + 1]);
  };
  while (tc--) {
    long long n;
    cin >> n;
    vector<pair<long long, long long>> val;
    for (long long i = 0, a; i < n; i++) {
      cin >> a;
      val.push_back({a, i});
    }
    sort(begin(val), end(val));
    vector<long long> permutation(n);
    long long inversion_cnt = 0;
    for (long long i = 0; i < n; i++) permutation[val[i].second] = i;
    for (long long i = 0; i < n; i++) {
      for (long long j = i + 1; j < n; j++) {
        if (permutation[j] < permutation[i]) inversion_cnt++;
      }
    }
    long long poss = 1;
    if (inversion_cnt & 1) {
      poss = 0;
      for (long long i = 0; i < n - 1; i++) {
        if (val[i].first == val[i + 1].first) {
          swap(permutation[val[i].second], permutation[val[i + 1].second]);
          poss = 1;
          break;
        }
      }
    }
    vector<long long> ret;
    if (!poss) {
      cout << "-1"
           << "\n";
      continue;
    }
    for (long long i = 0; i < n - 2; i++) {
      long long ind = min_element(begin(permutation) + i, end(permutation)) -
                      permutation.begin();
      while (ind > i + 1) {
        rotate(permutation, ind - 2);
        ret.push_back(ind - 2);
        ind -= 2;
      }
      if (ind != i) {
        rotate(permutation, i);
        ret.push_back(i);
        ret.push_back(i);
        rotate(permutation, i);
        ind = i;
      }
    }
    for (long long i = 0; i < 3; i++) {
      if (is_sorted(begin(permutation), end(permutation))) break;
      rotate(permutation, n - 3);
      ret.push_back(n - 3);
    }
    if (!is_sorted(begin(permutation), end(permutation)))
      cout << "-1"
           << "\n";
    else {
      cout << ret.size() << "\n";
      for (auto k : ret) cout << k + 1 << " ";
      cout << "\n";
    }
  }
}