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;
  void abc() { cerr << endl; }
  template <typename T, typename... U>
  void abc(T a, U... b) {
    cerr << a << ' ', abc(b...);
  }
  template <typename T>
  void printv(T l, T r) {
    while (l != r) cout << *l << " \n"[++l == r];
  }
  template <typename A, typename B>
  istream& operator>>(istream& o, pair<A, B>& a) {
    return o >> a.first >> a.second;
  }
  template <typename A, typename B>
  ostream& operator<<(ostream& o, pair<A, B> a) {
    return o << '(' << a.first << ", " << a.second << ')';
  }
  template <typename T>
  ostream& operator<<(ostream& o, vector<T> a) {
    bool is = false;
    for (T i : a) {
      o << (is ? ' ' : '{'), is = true, o << i;
    }
    return o << '}';
  }
  template <typename T>
  struct vv : vector<vector<T>> {
    vv(int n, int m, T v) : vector<vector<T>>(n, vector<T>(m, v)) {}
    vv() {}
  };
  template <typename T>
  struct vvv : vector<vv<T>> {
    vvv(int n, int m, int k, T v) : vector<vv<T>>(n, vv<T>(m, k, v)) {}
    vvv() {}
  };
  const int N = 200000, logN = 18, K = 1000;
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      vector<int> p(n);
      for (int i = 0; i < n; ++i) cin >> p[i];
      vector<int> ans;
      auto add = [&](int x) {
        ans.push_back(x);
        reverse(p.begin(), p.begin() + x);
      };
      bool is = true;
      for (int now = n; now > 1; now -= 2) {
        int f = -1, s = -1;
        for (int i = 0; i < now; ++i) {
          if (p[i] == now - 1) f = i;
          if (p[i] == now) s = i;
        }
        if (f & 1 && s & 1 ^ 1) {
          add(s + 1);
          for (int i = 0; i < now; ++i) {
            if (p[i] == now - 1) f = i;
            if (p[i] == now) s = i;
          }
          if (s + 1 != f) {
            add(f);
            for (int i = 0; i < now; ++i) {
              if (p[i] == now - 1) f = i;
              if (p[i] == now) s = i;
            }
          }
          add(now);
          for (int i = 0; i < now; ++i) {
            if (p[i] == now - 1) f = i;
            if (p[i] == now) s = i;
          }
          add(s + 1);
          add(now);
        } else {
          is = false;
          break;
        }
      }
      if (is) {
        void(0);
        cout << ans.size() << endl;
        printv(ans.begin(), ans.end());
      } else {
        cout << -1 << endl;
      }
    }
  }
  