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;
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) {
      int N;
      cin >> N;
      vector<int> a(N);
      bool good = true;
      for (int i = 0; i < N; ++i) {
        cin >> a[i];
        if (a[i] % 2 != (i + 1) % 2) {
          good = false;
        }
      }
      if (not good) {
        cout << "-1\n";
        continue;
      }
      int M = 0;
      vector<int> ps;
      auto apply = [&](int p) {
        assert(p % 2);
        ps.push_back(p);
        reverse(a.begin(), a.begin() + p);
        ++M;
      };
      for (int x = N; x > 1; x -= 2) {
        apply(int(find(a.begin(), a.end(), x) - a.begin()) + 1);
        int i = int(find(a.begin(), a.end(), x - 1) - a.begin());
        if (i > 1) {
          apply(i);
        }
        apply(i + 2);
        apply(3);
        apply(x);
      }
      assert(is_sorted(a.begin(), a.end()));
      cout << M << '\n';
      for (int j = 0; j < M; ++j) {
        cout << ps[j] << "\n "[j + 1 < M];
      }
    }
    exit(0);
  }
  