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;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int &t : a) cin >> t;
    vector<int> ans;
    auto apply_rot = [&](int id) {
      rotate(a.begin() + id, a.begin() + id + 2, a.begin() + id + 3);
      ans.push_back(id);
    };
    for (int i = 0; i + 2 < n; i++) {
      int p = min_element(a.begin() + i, a.end()) - a.begin();
      int v = a[p];
      for (; p >= i + 2; p -= 2) {
        apply_rot(p - 2);
      }
      while (a[i] != v) apply_rot(i);
    }
    if (a[n - 2] > a[n - 1]) {
      for (int i = n - 1; i > 0; i--) {
        if (a[i] == a[i - 1]) {
          int v = a[i];
          for (int j = i; j + 2 < n; j += 2) {
            while (a[j + 2] != v) apply_rot(j);
          }
          int p = n - 1;
          if (a[n - 1] != v)
            apply_rot(n - 3);
          else {
            apply_rot(n - 3);
            apply_rot(n - 3);
            p = n - 2;
          }
          for (; p >= i + 2; p -= 2) {
            apply_rot(p - 2);
          }
          if (p != i) {
            apply_rot(p - 2);
          }
          break;
        }
      }
    }
    if (a[n - 3] == a[n - 1]) apply_rot(n - 3);
    if (is_sorted(begin(a), end(a))) {
      cout << int(ans.size()) << endl;
      for (int t : ans) cout << t + 1 << " ";
      cout << endl;
    } else {
      puts("-1");
    }
  }
}