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;
  using ll = long long;
  using ld = long double;
  const ld eps = 1e-8;
  template <class U, class V>
  ostream& operator<<(ostream& out, const pair<U, V>& u) {
    return out << "(" << u.first << ", " << u.second << ")";
  }
  template <class Con, class = decltype(begin(declval<Con>()))>
  typename enable_if<!is_same<Con, string>::value, ostream&>::type operator<<(
      ostream& out, const Con& con) {
    out << "{";
    for (auto beg = con.begin(), it = beg; it != con.end(); ++it)
      out << (it == beg ? "" : ", ") << *it;
    return out << "}";
  }
  template <size_t i, class T>
  ostream& print_tuple_utils(ostream& out, const T& tup) {
    if constexpr (i == tuple_size<T>::value)
      return out << ")";
    else
      return print_tuple_utils<i + 1, T>(out << (i ? ", " : "(") << get<i>(tup),
                                         tup);
  }
  template <class... U>
  ostream& operator<<(ostream& out, const tuple<U...>& u) {
    return print_tuple_utils<0, tuple<U...>>(out, u);
  }
  const int maxn = 3000;
  int n;
  int a[maxn];
  int pos_of(int x) {
    int ans = 0;
    while (ans < n and a[ans] != x) ++ans;
    assert(ans < n);
    return ans;
  }
  vector<int> ans;
  void do_rev(int len) {
    assert(len & 1);
    ans.push_back(len);
    reverse(a, a + len);
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    int ntest;
    cin >> ntest;
    while (ntest--) {
      cin >> n;
      for (int i = (int)(0); i < (int)(n); ++i) cin >> a[i];
      ans.clear();
      bool ok = true;
      for (int len = n; len > 1; len -= 2) {
        int mx = len;
        int pmx = pos_of(mx);
        if (pmx % 2 == 1) {
          ok = false;
          break;
        }
        do_rev(pmx + 1);
        int pmx1 = pos_of(mx - 1);
        if (pmx1 % 2 == 0) {
          ok = false;
          break;
        }
        do_rev(pmx1);
        do_rev(pmx1 + 2);
        do_rev(3);
        do_rev(len);
      }
      if (!ok) {
        cout << "-1\n";
        continue;
      }
      cout << ((int)ans.size()) << '\n';
      for (auto i : ans) cout << i << ' ';
      cout << '\n';
      assert(is_sorted(a, a + n));
    }
    return 0;
  }
  