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 pii = std::pair<long long, long long>;
using namespace std;
const long long maxn = 505;
long long t, n, x;
pii a[maxn];
vector<long long> ops;
void rshift3(long long start) {
  assert(start >= 0 && start < n - 2);
  ops.push_back(start);
  pii temp = a[start + 2];
  a[start + 2] = a[start + 1];
  a[start + 1] = a[start];
  a[start] = temp;
}
long long isgood() {
  for (long long i = 0; i < n - 1; i++)
    if (a[i].first > a[i + 1].first) return 0;
  return 1;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> t;
  for (long long cases = 0; cases < t; cases++) {
    cin >> n;
    ops.clear();
    vector<pii> order;
    for (long long i = 0; i < n; i++) {
      cin >> x;
      a[i] = {x, i};
      order.push_back(a[i]);
    }
    sort(order.begin(), order.end());
    for (long long i = n - 1; i >= 2; i--) {
      long long curpos = -1;
      for (long long j = i; j >= 0; j--)
        if (a[j] == order[i]) {
          curpos = j;
          break;
        }
      assert(curpos != -1);
      while (curpos < i) {
        assert(a[curpos] == order[i]);
        if (curpos == 0)
          rshift3(0);
        else
          rshift3(curpos - 1);
        curpos++;
      }
    }
    for (long long i = 0; i < n - 2 && !isgood(); i++)
      if (a[i] != order[i])
        for (long long j = 0; j < 2 && a[i] != order[i]; j++) rshift3(i);
    if (!isgood())
      cout << "-1\n";
    else {
      cout << ops.size() << "\n";
      for (auto x : ops) cout << x + 1 << " ";
      cout << "\n";
    }
  }
  return 0;
}