Problem: You are given an array a consisting of n integers.

In one move, you can choose some index i (1 ≤ i ≤ n - 2) and shift the segment [a_i, a_{i + 1}, a_{i + 2}] cyclically to the right (i.e. replace the segment [a_i, a_{i + 1}, a_{i + 2}] with [a_{i + 2}, a_i, a_{i + 1}]). 

Your task is to sort the initial array by no more than n^2 such operations or say that it is impossible to do that.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (3 ≤ n ≤ 500) — the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 500), where a_i is the i-th element a.

It is guaranteed that the sum of n does not exceed 500.

Output

For each test case, print the answer: -1 on the only line if it is impossible to sort the given array using operations described in the problem statement, or the number of operations ans on the first line and ans integers idx_1, idx_2, ..., idx_{ans} (1 ≤ idx_i ≤ n - 2), where idx_i is the index of left border of the segment for the i-th operation. You should print indices in order of performing operations.

Example

Input


5
5
1 2 3 4 5
5
5 4 3 2 1
8
8 4 5 2 3 6 7 3
7
5 2 1 6 4 7 3
6
1 2 3 3 6 4


Output


0

6
3 1 3 2 2 3 
13
2 1 1 6 4 2 4 3 3 4 4 6 6 
-1
4
3 3 4 4 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
mt19937 gen(time(0));
bool IS_ONE_TEST = 0;
vector<int> res;
int total = 0;
bool proc(int idx, vector<int> &a) {
  if (!total) return 0;
  res.push_back(idx + 1);
  int temp = a[idx + 2];
  a[idx + 2] = a[idx + 1];
  a[idx + 1] = a[idx];
  a[idx] = temp;
  --total;
  return 1;
}
void solve() {
  int n;
  cin >> n;
  res.clear();
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<int> b = a;
  total = n * n;
  sort(a.begin(), a.end());
  for (int z = 0; z < n; ++z) {
    for (int i = 0; i + 2 < n; ++i) {
      int mn = min({b[i], b[i + 1], b[i + 2]});
      while (b[i] != mn && proc(i, b))
        ;
    }
    for (int i = n - 1; i > 1; --i) {
      int mx = max({b[i], b[i - 1], b[i - 2]});
      while (b[i] != mx && proc(i - 2, b))
        ;
    }
  }
  if (a == b) {
    cout << res.size() << endl;
    for (auto &c : res) cout << c << " ";
    cout << endl;
  } else
    cout << -1 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(15);
  int t;
  if (IS_ONE_TEST)
    t = 1;
  else
    cin >> t;
  while (t-- > 0) {
    solve();
  }
}