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 int NMAX = 500;
int N;
vector<pair<int, int> > v;
int a[NMAX + 5], pos[NMAX + 5];
void Test() {
  cin >> N;
  v.clear();
  for (int i = 1; i <= N; i++) {
    int x;
    cin >> x;
    v.push_back({x, i});
  }
  bool duplicate = false;
  int swapVal;
  sort(v.begin(), v.end());
  for (int i = 1; i < (int)v.size(); i++)
    if (v[i].first == v[i - 1].first) {
      swapVal = i;
      duplicate = true;
    }
  int k = 0;
  for (auto it : v) a[it.second] = ++k;
  int nrInv = 0;
  for (int i = 1; i < N; i++)
    for (int j = i + 1; j <= N; j++)
      if (a[i] > a[j]) nrInv++;
  if (nrInv % 2 == 1) {
    if (duplicate == false) {
      cout << -1 << '\n';
      return;
    }
    int p1 = -1, p2 = -1;
    for (int i = 1; i <= N; i++) {
      if (a[i] == swapVal)
        p1 = i;
      else if (a[i] == swapVal + 1)
        p2 = i;
    }
    swap(a[p1], a[p2]);
  }
  for (int i = 1; i <= N; i++) pos[a[i]] = i;
  vector<int> sol;
  for (int i = 1; i <= N - 2; i++) {
    while (a[i] != i) {
      if (pos[i] >= i + 2) {
        sol.push_back(pos[i] - 2);
        int a1 = a[pos[i] - 2];
        int a2 = a[pos[i] - 1];
        int a3 = a[pos[i]];
        a[pos[i] - 2] = a3;
        a[pos[i] - 1] = a1;
        a[pos[i]] = a2;
        pos[a1] = pos[i] - 1;
        pos[a2] = pos[i];
        pos[a3] = pos[i] - 2;
      } else {
        sol.push_back(i);
        int a1 = a[i];
        int a2 = a[i + 1];
        int a3 = a[i + 2];
        a[i] = a3;
        a[i + 1] = a1;
        a[i + 2] = a2;
        pos[a1] = i + 1;
        pos[a2] = i + 2;
        pos[a3] = i;
      }
    }
  }
  if (a[N - 1] == N - 1 && a[N] == N) {
    cout << (int)sol.size() << '\n';
    for (auto it : sol) cout << it << ' ';
    cout << '\n';
  } else {
    cout << -1 << '\n';
    return;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) Test();
  return 0;
}