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;
const long long mod = 1000000007;
const int inf = 1e9 + 10;
const long long INF = 1e18;
const long double EPS = 1e-10;
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(25);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < (n); ++i) cin >> a[i];
    vector<pair<int, int>> p(n);
    for (int i = 0; i < (n); ++i) p[i] = {a[i], i};
    sort(p.begin(), p.end());
    int cnt = 0;
    for (int i = 0; i < (n); ++i)
      for (int j = 0; j < (i); ++j)
        if (p[j].second > p[i].second) cnt++;
    for (int i = 0; i < (n - 1); ++i) {
      if (cnt & 1 && p[i].first == p[i + 1].first) {
        swap(p[i], p[i + 1]);
        cnt--;
        break;
      }
    }
    if (cnt & 1) {
      cout << -1 << '\n';
      continue;
    }
    for (int i = 0; i < (n); ++i) a[p[i].second] = i;
    vector<int> ans;
    for (int i = 0; i < (n - 2); ++i) {
      int idx;
      for (int j = (n - 1); j >= 0; --j) {
        if (a[j] == i) {
          idx = j;
          break;
        }
      }
      while (idx - i >= 2) {
        ans.push_back(idx - 2);
        a[idx] = a[idx - 1];
        a[idx - 1] = a[idx - 2];
        a[idx - 2] = i;
        idx -= 2;
      }
      if (idx - i == 1) {
        ans.push_back(idx - 1);
        ans.push_back(idx - 1);
        a[idx] = a[idx + 1];
        a[idx + 1] = a[idx - 1];
        a[idx - 1] = i;
        idx--;
      }
      assert(idx == i);
    }
    cout << ans.size() << '\n';
    for (int x : ans) cout << x + 1 << ' ';
    cout << '\n';
  }
  return 0;
}