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
#include <bits/stdc++.h>
using namespace std;
int n, a[(int)501], ind[(int)501], rep;
set<int> aux;
auto cmp = [](int i, int j) { return a[i] < a[j]; };
void cycle(int i) {
  swap(ind[a[i]], ind[a[i + 1]]);
  swap(a[i], a[i + 1]);
  swap(ind[a[i]], ind[a[i + 2]]);
  swap(a[i], a[i + 2]);
}
void solve() {
  aux.clear();
  rep = -1;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i], ind[i] = i;
    if (!aux.insert(a[i]).second) rep = a[i];
  }
  sort(ind, ind + n, cmp);
  int inversions = 0;
  for (int i = 0; i < n; ++i)
    for (int j = i + 1; j < n; ++j)
      if (ind[i] > ind[j]) inversions++;
  if (inversions % 2 && rep == -1) {
    cout << -1 << endl;
    return;
  }
  if (inversions % 2)
    for (int i = 0; i < n; ++i)
      if (a[ind[i]] == rep) {
        swap(ind[i], ind[i + 1]);
        break;
      }
  for (int i = 0; i < n; ++i) a[ind[i]] = i;
  vector<int> ans;
  for (int i = n - 1; i >= 2; --i)
    while (ind[i] < i) {
      ans.push_back(min(i - 2, ind[i]));
      cycle(ans.back());
    }
  cout << ans.size() << endl;
  for (int x : ans) cout << x + 1 << " ";
  cout << endl;
}
int main() {
  int T;
  cin >> T;
  while (T--) solve();
}
