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;
void shikhar7s(long long int cas) {
  long long int n, i;
  cin >> n;
  long long int a[n];
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int j, in, mi;
  vector<long long int> ans;
  for (i = 0; i < n - 2; i++) {
    mi = 2000000000000000000;
    for (j = i; j < n; j++) {
      if (a[j] < mi) {
        mi = a[j];
        in = j;
      }
    }
    while (in - i > 1) {
      ans.push_back(in - 2);
      mi = a[in];
      a[in] = a[in - 1];
      a[in - 1] = a[in - 2];
      a[in - 2] = mi;
      in -= 2;
    }
    if (in != i) {
      ans.push_back(i);
      ans.push_back(i);
      mi = a[in];
      a[in] = a[in + 1];
      a[in + 1] = a[in - 1];
      a[in - 1] = mi;
    }
  }
  if (a[n - 2] > a[n - 1]) {
    if (a[n - 1] == a[n - 3]) {
      ans.push_back(n - 3);
    } else {
      for (i = n - 2; i >= 0; i--) {
        if (a[i] == a[i + 1]) break;
      }
      if (i == -1) {
        cout << -1 << "\n";
        return;
      }
      ans.push_back(i + 1);
      ans.push_back(i);
      i += 2;
      mi = a[i];
      a[i] = a[i + 1];
      a[i + 1] = mi;
      for (i = i; i < n - 2; i++) {
        mi = 2000000000000000000;
        for (j = i; j < n; j++) {
          if (a[j] < mi) {
            mi = a[j];
            in = j;
          }
        }
        while (in - i > 1) {
          ans.push_back(in - 2);
          mi = a[in];
          a[in] = a[in - 1];
          a[in - 1] = a[in - 2];
          a[in - 2] = mi;
          in -= 2;
        }
        if (in != i) {
          ans.push_back(in - 1);
          ans.push_back(in - 1);
          mi = a[in];
          a[in] = a[in + 1];
          a[in + 1] = a[in - 1];
          a[in - 1] = mi;
        }
      }
    }
  }
  cout << (long long int)ans.size() << "\n";
  for (i = 0; i < (long long int)ans.size(); i++) {
    cout << ans[i] + 1 << " ";
  }
  cout << "\n";
}
signed main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  cin >> t;
  long long int cas = 1;
  while (cas <= t) {
    shikhar7s(cas);
    cas++;
  }
  return 0;
}