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;
long long a[501];
vector<long long> ans;
inline void change(long long x) {
  ans.push_back(x);
  swap(a[x + 1], a[x + 2]);
  swap(a[x], a[x + 1]);
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    ans.clear();
    vector<pair<long long, long long> > lis;
    cin >> n;
    for (long long i = 1; i <= n; ++i) {
      cin >> a[i];
      lis.push_back(make_pair(a[i], i));
    }
    sort((lis).begin(), (lis).end());
    for (long long i = 1; i <= n; ++i) a[lis[i - 1].second] = i;
    long long cnt = 0;
    for (long long i = 1; i <= n; ++i)
      for (long long j = i + 1; j <= n; ++j) cnt += (a[i] > a[j]);
    if (cnt % 2) {
      for (long long i = 1; i <= n; ++i)
        if (lis[i - 1].first == lis[i].first) {
          swap(a[lis[i - 1].second], a[lis[i].second]);
          break;
        }
    }
    for (long long i = 1; i <= n - 2; ++i) {
      long long pos = min_element(a + i, a + n + 1) - a;
      while (pos - 2 >= i) {
        change(pos - 2);
        pos -= 2;
      }
      if (pos == i + 1) {
        change(i);
        change(i);
      }
    }
    for (long long i = 0; i < 3; ++i)
      if (!is_sorted(a + 1, a + n + 1)) change(n - 2);
    if (!is_sorted(a + 1, a + n + 1))
      cout << -1 << '\n';
    else {
      cout << (long long)(ans).size() << '\n';
      for (auto i : ans) cout << i << " ";
      cout << '\n';
    }
  }
}