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 mod = 1e9 + 7, INF = 1e18;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  unsigned long long t = 1;
  cin >> t;
  while (t--) {
    int n, flag = 0, flag2 = 0;
    cin >> n;
    vector<int> a(n), ans, b;
    for (int i = 0; i < n; i++) cin >> a[i];
    b = a;
    sort(b.begin(), b.end());
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) {
        int p = j;
        for (int k = j; k < n; k++)
          if (a[k] == b[j]) {
            p = k;
            break;
          }
        while (p != j) {
          if (p - 2 >= j) {
            int temp = a[p];
            a[p] = a[p - 1];
            a[p - 1] = a[p - 2];
            a[p - 2] = temp;
            p -= 2;
            ans.push_back(p + 1);
          } else if (p < n - 1) {
            int temp = a[p];
            a[p] = a[p + 1];
            a[p + 1] = a[p - 1];
            a[p - 1] = temp;
            p -= 1;
            ans.push_back(p + 1);
            ans.push_back(p + 1);
          } else
            break;
        }
      }
    for (int i = 0; i < n - 1; i++) {
      if (b[i] == b[i + 1]) flag2 = i + 1;
      if (a[i] > a[i + 1]) flag = 1;
    }
    if (flag && flag2 == 0) {
      cout << "-1\n";
      continue;
    }
    if (flag) {
      for (int i = n - 2; i >= flag2; i--) {
        ans.push_back(i);
        int temp = a[i];
        a[i] = a[i - 1];
        a[i - 1] = a[i + 1];
        a[i + 1] = temp;
      }
    }
    cout << ans.size() << "\n";
    for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
    cout << "\n";
  }
  return 0;
}