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>
  #pragma GCC target("avx2")
  #pragma GCC optimize("unroll-loops")
  #pragma GCC optimize("Ofast")
  constexpr long long INF = 200000000000000001LL;
  constexpr int INF32 = 2100000001;
  size_t seed42 = std::chrono::system_clock::now().time_since_epoch().count();
  std::mt19937 rng(seed42);
  std::uniform_int_distribution<int> unidistrib;
  int randint() { return unidistrib(rng); }
  using namespace std;
  unsigned long long modulo = 1000000007;
  int modulo32 = 1000000007;
  unsigned long long poww(unsigned long long x, unsigned long long n) {
    if (n == 0) return 1;
    unsigned long long answ = poww(x, n / 2);
    answ = (answ * answ) % modulo;
    if (n % 2) answ = (answ * x) % modulo;
    return answ;
  }
  pair<int, int> operator+(const pair<int, int>& lhs, pair<int, int>& rhs) {
    return {lhs.first + rhs.first, lhs.second + rhs.second};
  }
  template <class T>
  ostream& operator<<(ostream& lhs, pair<T, T>& rhs) {
    return (lhs << rhs.first << ":" << rhs.second);
  }
  int main() {
    ios::sync_with_stdio(false);
    std::cin.tie(0);
    int t, n, k, m;
    cin >> t;
    while (t--) {
      cin >> n;
      vector<int> a(n);
      vector<int> a0, a1;
      for (int i = 0; i < n; ++i) {
        cin >> a[i];
        if (i % 2)
          a1.push_back(a[i]);
        else
          a0.push_back(a[i]);
      }
      auto a_sorted = a;
      sort(a_sorted.begin(), a_sorted.end());
      sort(a0.begin(), a0.end());
      sort(a1.begin(), a1.end());
      bool good = true;
      for (int i = 0; i < n; ++i) {
        int tmp = i % 2 ? a1[i / 2] : a0[i / 2];
        if (a_sorted[i] != tmp) {
          good = false;
          break;
        }
      }
      if (!good) {
        cout << "-1\n";
        continue;
      }
      reverse(a0.begin(), a0.end());
      reverse(a1.begin(), a1.end());
      vector<int> answ;
      for (int i = 0; i < n / 2; ++i) {
        auto it = find(a.begin(), a.end(), a0[i]);
        if (it != a.begin()) {
          answ.push_back((it - a.begin()) + 1);
          reverse(a.begin(), it + 1);
        }
        it = find(a.begin(), a.end(), a1[i]);
        if (it != a.begin() + 1) {
          answ.push_back(it - a.begin());
          reverse(a.begin(), it);
        }
      }
      int minn = a_sorted[0];
      for (int i = 0; i < n / 2; ++i) {
        auto it = find(a.begin(), a.end(), minn);
        if (it != a.begin()) {
          answ.push_back((it - a.begin()) + 1);
          reverse(a.begin(), it + 1);
        }
        it = find(a.begin(), a.end(), a0[i]);
        answ.push_back((it - a.begin()) + 1);
        reverse(a.begin(), it + 1);
        answ.push_back(n - 2 * i);
        reverse(a.begin(), a.end() - 2 * i);
      }
      cout << answ.size() << "\n";
      for (auto el : answ) {
        cout << el << " ";
      };
      cout << "\n";
    }
    return 0;
  }
  