description: You have a permutation: an array a = [a_1, a_2, …, a_n] of distinct integers from 1 to n. The length of the permutation n is odd.
  
  You need to sort the permutation in increasing order.
  
  In one step, you can choose any prefix of the permutation with an odd length and reverse it. Formally, if a = [a_1, a_2, …, a_n], you can choose any odd integer p between 1 and n, inclusive, and set a to [a_p, a_{p-1}, …, a_1, a_{p+1}, a_{p+2}, …, a_n].
  
  Find a way to sort a using no more than 5n/2 reversals of the above kind, or determine that such a way doesn't exist. The number of reversals doesn't have to be minimized.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
  
  The first line of each test case contains a single integer n (3 ≤ n ≤ 2021; n is odd) — the length of the permutation.
  
  The second line contains n distinct integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — the permutation itself. 
  
  It is guaranteed that the sum of n over all test cases does not exceed 2021.
  
  Output
  
  For each test case, if it's impossible to sort the given permutation in at most 5n/2 reversals, print a single integer -1.
  
  Otherwise, print an integer m (0 ≤ m ≤ 5n/2), denoting the number of reversals in your sequence of steps, followed by m integers p_i (1 ≤ p_i ≤ n; p_i is odd), denoting the lengths of the prefixes of a to be reversed, in chronological order.
  
  Note that m doesn't have to be minimized. If there are multiple answers, print any.
  
  Example
  
  Input
  
  
  3
  3
  1 2 3
  5
  3 4 5 2 1
  3
  2 1 3
  
  
  Output
  
  
  4
  3 3 3 3
  2
  3 5
  -1
  
  Note
  
  In the first test case, the permutation is already sorted. Any even number of reversals of the length 3 prefix doesn't change that fact.
  
  In the second test case, after reversing the prefix of length 3 the permutation will change to [5, 4, 3, 2, 1], and then after reversing the prefix of length 5 the permutation will change to [1, 2, 3, 4, 5].
  
  In the third test case, it's impossible to sort the permutation.
solution: #include <bits/stdc++.h>
  using namespace std;
  template <class T>
  ostream& operator<<(ostream& out, vector<T>& v) {
    out << "[";
    for (auto k : v) out << k << " ";
    out << "]"
        << "\n";
    return out;
  }
  template <class T>
  ostream& operator<<(ostream& out, set<T> s) {
    out << "{";
    for (auto k : s) out << k << " ";
    out << "}"
        << "\n";
    return out;
  }
  template <class T, class U>
  ostream& operator<<(ostream& out, pair<T, U> p) {
    out << "[ " << p.first << " , " << p.second << " ] ";
    return out;
  }
  template <class T, class U>
  istream& operator>>(istream& in, pair<T, U>& p) {
    in >> p.first >> p.second;
    return in;
  }
  void Caso() {
    int n;
    cin >> n;
    vector<int> in(n);
    for (int i = 0; i < int(n); i++) cin >> in[i];
    vector<int> res;
    int np = n - (1 - n % 2);
    for (int i = 0; i < int(n); i++)
      if (in[i] % 2 == i % 2) {
        cout << -1 << '\n';
        return;
      }
    while (np > 1) {
      int y = np - 1;
      int y2 = 0;
      while (in[y2] != y) y2++;
      int x = 0;
      while (in[x] != np) x++;
      if (y2 == n - 1) {
        cout << -1 << '\n';
        return;
      }
      if (y2 != x - 1) {
        res.push_back(x + 1);
        reverse(in.begin(), in.begin() + x + 1);
        ;
        if (y2 < x) y2 = x - y2;
        x = 0;
        res.push_back(y2);
        reverse(in.begin(), in.begin() + y2);
        ;
        x = y2 - 1;
        res.push_back(y2 + 2);
        reverse(in.begin(), in.begin() + y2 + 2);
        ;
        x = 2;
        y2 = 1;
      }
      if (x + 1 == np) {
        np -= 2;
        continue;
      }
      if (x % 2 == 1) {
        cout << -1 << '\n';
        return;
      }
      if (x) res.push_back(x + 1);
      reverse(in.begin(), in.begin() + x + 1);
      ;
      res.push_back(np);
      reverse(in.begin(), in.begin() + np);
      ;
      np -= 2;
    }
    for (int i = 0; i < int(n - 1); i++)
      if (in[i] > in[i + 1]) {
        cout << -1 << '\n';
        return;
      }
    cout << res.size() << '\n';
    for (auto x : res) cout << x << " ";
    cout << '\n';
  }
  int main() {
    cin.tie(0);
    cin.sync_with_stdio(0);
    int t;
    cin >> t;
    while (t--) Caso();
    return 0;
  }
  