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;
  const int64_t inf = 1000000000000000000;
  const int64_t N = 4e6 + 5;
  const int64_t p = 1e9 + 7;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int q;
    cin >> q;
    while (q--) {
      int n;
      cin >> n;
      std::vector<int> a(n);
      for (int i = 0; i < n; ++i) cin >> a[i];
      int Z = 0;
      for (int i = 0; i < n; ++i)
        if (a[i] % 2 == i % 2) Z = 1;
      if (Z) {
        cout << -1 << '\n';
        continue;
      }
      std::vector<int> R;
      int i = 1, k = n;
      while (i < n) {
        for (int j = 0; j < k; ++j)
          if (a[j] == i) {
            R.push_back(j + 1);
            for (int l = 0; l < j / 2; ++l) swap(a[l], a[j - l]);
            break;
          }
        for (int j = 0; j < k; ++j)
          if (a[j] == i + 1) {
            R.push_back(j);
            for (int l = 0; l < (j - 1) / 2; ++l) swap(a[l], a[j - 1 - l]);
            break;
          }
        for (int j = 0; j < k; ++j)
          if (a[j] == i + 1) {
            R.push_back(j + 2);
            for (int l = 0; l < (j + 1) / 2; ++l) swap(a[l], a[j + 1 - l]);
            break;
          }
        {
          R.push_back(3);
          swap(a[0], a[2]);
        }
        {
          R.push_back(k);
          for (int i = 0; i < k / 2; ++i) swap(a[i], a[k - i - 1]);
        }
        k -= 2;
        i += 2;
      }
      R.push_back(n);
      cout << int(R.size()) << '\n';
      for (int i = 0; i < int(R.size()); ++i) cout << R[i] << ' ';
      cout << '\n';
    }
  }
  