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;
mt19937 mrand(43);
const double PI = acos((double)-1);
const double eps = 1e-5;
const long long inf0 = 1023 * 1024 * 1024;
const long long inf = inf0 * inf0;
const long long mod = 998244353;
void solve();
void scan();
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  cout << fixed;
  cout.precision(15);
  solve();
  return 0;
}
const int N = 501;
vector<int> id[N];
int a[N];
int p[N];
vector<int> ans;
void sw(int i) {
  ans.push_back(i);
  int x = a[i], y = a[i + 1], z = a[i + 2];
  a[i] = z, a[i + 1] = x, a[i + 2] = y;
  x = p[i], y = p[i + 1], z = p[i + 2];
  p[i] = z, p[i + 1] = x, p[i + 2] = y;
}
void solve() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    fill(a, a + n, 0);
    fill(p, p + n, 0);
    ans.clear();
    for (int i = 0; i <= n; i++) id[i].clear();
    vector<int> b(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      b[i] = a[i];
      id[a[i]].push_back(i);
    }
    sort((b).begin(), (b).end());
    for (int i = 0; i < N; i++) sort((id[i]).rbegin(), (id[i]).rend());
    for (int i = 0; i < n; i++) {
      p[id[b[i]].back()] = i;
      id[b[i]].pop_back();
    }
    for (int i = 0; i < (n - 2); i++) {
      int x = 0;
      for (int j = 0; j < n; j++) {
        if (p[j] == i) {
          x = j;
          break;
        }
      }
      while (x != i) {
        if (x - 1 == i) {
          sw(i);
          sw(i);
          x--;
        } else {
          sw(x - 2);
          x -= 2;
        }
      }
    }
    if (a[n - 1] < a[n - 2]) {
      if (a[n - 3] == a[n - 1]) {
        sw(n - 3);
      } else {
        int i = -1;
        for (int j = 0; j < n - 1; j++) {
          if (a[j] == a[j + 1]) {
            i = j;
            break;
          }
        }
        if (i == -1) {
          cout << -1 << '\n';
          continue;
        }
        while (true) {
          if (i == n - 2) {
            break;
          }
          if (i == n - 3) {
            sw(i);
            sw(i);
            break;
          }
          sw(i + 1);
          sw(i);
          i += 2;
        }
      }
    }
    cout << ans.size() << '\n';
    for (int i : ans) cout << i + 1 << " ";
    cout << '\n';
  }
  return;
}