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;
using namespace std::chrono;
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
template <typename A, typename B>
istream &operator>>(istream &cin, pair<A, B> &p) {
  cin >> p.first;
  return cin >> p.second;
}
mt19937 rng(61378913);
void usaco(string filename) {
  freopen((filename + ".in").c_str(), "r", stdin);
  freopen((filename + ".out").c_str(), "w", stdout);
}
const long double pi = 3.14159265358979323846;
const long long mod = 1000000007;
long long n, m, k, q, l, r, x, y, z;
long long a[1000005];
long long b[1000005];
long long c[1000005];
string second, t;
long long ans = 0;
void cyc(long long pos) {
  b[pos + 1] = a[pos];
  b[pos + 2] = a[pos + 1];
  b[pos] = a[pos + 2];
  for (long long i = (pos); i < (pos + 3); ++i) a[i] = b[i];
}
void solve(int tc) {
  cin >> n;
  for (int ele = 0; ele < n; ele++) cin >> a[ele];
  ;
  vector<long long> ops;
  for (long long i = 0; i < (n + 5); ++i) {
    if (i == n + 1) {
      long long p = -1;
      for (long long i = (3); i < (n - 1); ++i)
        if (a[i] == a[i + 1]) {
          p = i;
          break;
        }
      if (p != -1) {
        --p;
        while (p >= 3) {
          for (long long i = 0; i < (2); ++i) {
            ops.push_back(p + 1);
            cyc(p);
          }
          --p;
        }
        for (long long i = 0; i < (2); ++i) {
          ops.push_back(2);
          cyc(1);
        }
        for (long long i = 0; i < (2); ++i) {
          ops.push_back(3);
          cyc(2);
        }
        p = 3;
        while (p + 2 < n) {
          if (a[p + 1] <= a[p + 2]) break;
          ops.push_back(p + 1);
          cyc(p);
          ++p;
        }
      } else {
        if (n > 3) {
          for (long long i = 0; i < (3); ++i)
            if (a[i] == a[3]) {
              for (long long j = 0; j < (2 - i); ++j) {
                ops.push_back(1);
                cyc(0);
              }
              for (long long i = 0; i < (2); ++i) {
                ops.push_back(1);
                cyc(0);
              }
              for (long long i = 0; i < (2); ++i) {
                ops.push_back(2);
                cyc(1);
              }
              break;
            }
        }
      }
    }
    long long wpos = -1;
    for (long long j = (n - 2); j >= 0; --j) {
      if (a[j] > a[j + 1]) {
        if (wpos == -1 || a[j] > a[wpos]) wpos = j;
      }
    }
    if (wpos == -1) break;
    long long st = wpos - 1;
    if (st == -1) {
      ops.push_back(1);
      cyc(0);
      st = 0;
    }
    while (st + 1 < n - 1) {
      if (a[st + 1] <= a[st + 2]) break;
      ops.push_back(st + 1);
      cyc(st);
      ++st;
    }
  }
  if (((long long)ops.size()) > n * n) {
    cout << -1 << '\n';
    return;
  }
  bool is = 1;
  for (long long i = 0; i < (n - 1); ++i)
    if (a[i] > a[i + 1]) is = 0;
  if (!is) {
    cout << -1 << '\n';
  } else {
    cout << ops.size() << '\n';
    {
      for (int ele = 0; ele < (ops.size()); ele++) {
        if (ele) cout << " ";
        cout << ops[ele];
      }
      cout << '\n';
    };
  }
}
int main() {
  { ios_base::sync_with_stdio(false); }
  {
    cin.tie(NULL);
    cout.tie(NULL);
  }
  int tc = 1;
  cin >> tc;
  for (int t = 0; t < tc; t++) solve(t);
}