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;
using LL = long long;
const int N = 2e6 + 5;
int a[N];
pair<int, int> b[N];
int main() {
  ios::sync_with_stdio(false);
  int _;
  for (cin >> _; _; _--) {
    int inv = 0, n;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i], b[i] = {a[i], i};
    sort(b + 1, b + 1 + n);
    for (int i = 1; i <= n; i++) a[b[i].second] = i;
    for (int i = 1; i <= n; i++)
      for (int j = i + 1; j <= n; j++)
        if (a[i] > a[j]) inv++;
    if (inv & 1) {
      for (int i = 1; i < n; i++)
        if (b[i].first == b[i + 1].first) {
          swap(a[b[i].second], a[b[i + 1].second]);
          break;
        }
    }
    vector<int> ans;
    for (int i = 1; i <= n; i++) {
      int st = i;
      for (int j = n; j >= st + 2; j--) {
        if (a[j] < a[j - 1] && a[j] < a[j - 2]) {
          swap(a[j], a[j - 2]), swap(a[j], a[j - 1]);
          ans.push_back(j - 2);
        }
      }
      if (st + 2 <= n) {
        while (a[st + 1] < a[st] || a[st + 2] < a[st]) {
          swap(a[st + 2], a[st]), swap(a[st + 2], a[st + 1]);
          ans.push_back(st);
        }
      }
    }
    int suc = 1;
    for (int i = 2; i <= n; i++)
      if (a[i] <= a[i - 1]) suc = 0;
    if (suc) {
      cout << ans.size() << endl;
      for (auto v : ans) cout << v << " ";
      cout << endl;
    } else
      cout << "-1\n";
  }
  return 0;
}