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
#include <bits/stdc++.h>
using namespace std;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
pair<int, int> p[501];
int a[502];
int cnt_inv(int n) {
  int c = 0;
  for (int i = 0; i < (int)(n); ++i) {
    for (int j = i + 1; j < n; j++)
      if (a[i] > a[j]) c++;
  }
  return c;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 0; i < (int)(n); ++i) {
      cin >> p[i].first;
      p[i].second = i;
    }
    sort(p, p + n);
    int j = -1, te = p[0].first, f = 0;
    for (int i = 0; i < (int)(n); ++i) {
      a[p[i].second] = i + 1;
    }
    for (int i = 0; i < (int)(n); ++i) {
      if (p[i].first == te)
        f++;
      else {
        f = 1;
        te = p[i].first;
      }
      if (f >= 2) {
        j = te;
        break;
      }
    }
    int inv = cnt_inv(n);
    if (inv % 2 != 0) {
      if (j == -1) {
        cout << "-1"
             << "\n";
        continue;
      } else {
        int ind = 0;
        while (p[ind].first != j) ind++;
        a[p[ind].second] = ind + 2;
        a[p[ind + 1].second] = ind + 1;
      }
    }
    vector<int> v;
    for (int i = 0; i < (int)(n); ++i) {
      int k = 0;
      while (a[k] != i + 1) k++;
      while (k > (i + 1)) {
        v.push_back(k - 2 + 1);
        swap(a[k - 1], a[k - 2]);
        swap(a[k - 2], a[k]);
        k -= 2;
      }
      if (k == i) continue;
      v.push_back(i + 1);
      v.push_back(i + 1);
      swap(a[i + 1], a[i + 2]);
      swap(a[i + 2], a[i]);
    }
    cout << v.size() << "\n";
    for (auto x : v) cout << x << " ";
    cout << "\n";
  }
}
