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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int tt = 1;
  cin >> tt;
  for (long long int tc = 1; tc <= tt; tc++) {
    long long int n;
    cin >> n;
    vector<long long int> arr(n), brr(n);
    for (long long int i = 0; i <= n - 1; i++) {
      cin >> arr[i];
      brr[i] = arr[i];
    }
    sort(brr.begin(), brr.end());
    vector<long long int> ans;
    bool bl = 1;
    for (long long int i = 0; i <= n - 3; i++) {
      if (arr[i] == brr[i]) continue;
      long long int index = -1, index2 = -1;
      for (long long int j = i; j <= n - 1; j++) {
        if (arr[j] == brr[i] and (j - i) % 2 == 0) {
          index = j;
          break;
        }
      }
      for (long long int j = i; j <= n - 1; j++) {
        if (arr[j] == brr[i] and (j - i) % 2 == 1) {
          index2 = j;
          break;
        }
      }
      if (index == -1) index = index2;
      long long int j = index - 2;
      while (j >= i) {
        ans.push_back(j);
        j -= 2;
      }
      j += 2;
      for (long long int k = index; k >= j + 1; k--) {
        arr[k] = arr[k - 1];
      }
      if (j != i) {
        ans.push_back(i);
        ans.push_back(i);
        arr[i + 1] = arr[i + 2];
        arr[i + 2] = arr[i];
      }
      arr[i] = brr[i];
    }
    if (arr[n - 1] != brr[n - 1]) bl = 0;
    if (!bl) {
      if (arr[n - 3] == arr[n - 1]) {
        bl = 1;
        swap(arr[n - 1], arr[n - 2]);
        ans.push_back(n - 3);
      }
    }
    if (!bl) {
      for (long long int i = 0; i <= n - 4; i++) {
        if (arr[i] == arr[i + 1]) {
          bl = 1;
          swap(arr[n - 1], arr[n - 2]);
          long long int j = i;
          while (j < n - 4) {
            ans.push_back(j);
            j++;
          }
          ans.push_back(n - 3);
          ans.push_back(n - 4);
          j = n - 5;
          while (j >= i) {
            ans.push_back(j);
            ans.push_back(j);
            j--;
          }
          break;
        }
      }
    }
    if (bl and ans.size() <= n * n) {
      cout << ans.size();
      cout << '\n';
      for (auto x : ans) cout << x + 1 << " ";
    } else
      cout << "-1";
    cout << '\n';
  }
  return 0;
}