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>
#pragma GCC target("avx2")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
T sqr(T a) {
  return a * a;
}
template <typename T>
int sign(T a) {
  if (a == 0) return 0;
  return (a > 0 ? 1 : -1);
}
template <typename T>
bool uax(T& a, const T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool uin(T& a, const T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2>& p) {
  out << p.first << ' ' << p.second;
  return out;
}
mt19937_64 rnd(0);
const int LOG = 64;
const int N = (int)2e5 + 7;
const int MAXN = (int)1e6 + 7;
const int MOD = (int)1e9 + 7;
const int INF = (int)2e9 + 7;
const int CONST = 450;
const long long LINF = (int64_t)1e18;
const long double PI = 3.1415926535897932384626433832795;
const long double EPS = 1;
vector<int> act;
int n, a[N], need[N];
bool check_parity() {
  int inv = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) inv += need[i] > need[j];
  }
  return inv % 2 == 0;
}
void perform(int pos) {
  act.push_back(pos + 1);
  swap(need[pos], need[pos + 1]);
  swap(need[pos], need[pos + 2]);
}
void solve() {
  cin >> n;
  act.clear();
  vector<pair<int, int>> v;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    v.push_back(make_pair(a[i], i));
  }
  sort(v.begin(), v.end());
  pair<int, int> eq = make_pair(-1, -1);
  for (int i = 0; i < n; i++) {
    need[v[i].second] = i;
    if (i && v[i - 1].first == v[i].first)
      eq = make_pair(v[i - 1].second, v[i].second);
  }
  if (!check_parity()) {
    if (eq.first == -1) {
      cout << "-1\n";
      return;
    }
    swap(need[eq.first], need[eq.second]);
  }
  for (int i = 0; i < n; i++) {
    int pos;
    for (int j = 0; j < n; j++) {
      if (need[j] == i) pos = j;
    }
    if (pos % 2 != i % 2) {
      if (pos == n - 1) {
        perform(n - 3);
        perform(n - 3);
        pos = n - 2;
      } else {
        perform(pos - 1);
        pos++;
      }
    }
    while (pos != i) {
      perform(pos - 2);
      pos -= 2;
    }
  }
  cout << (int)act.size() << "\n";
  for (int x : act) cout << x << ' ';
  cout << "\n";
}
int main() {
  double start = clock();
  cout << fixed << setprecision(20);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  rnd.seed(time(0));
  int t = 1;
  cin >> t;
  for (int i = 0; i < t; i++) {
    solve();
  }
}