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 long long INF = 1e9;
  const long long MOD = 1e9 + 7;
  const long long MX = 1e6;
  const long double PI = 3.141592654;
  int a[2022];
  void solve(int r) {
    int l = 1;
    while (l < r) {
      swap(a[l], a[r]);
      l++;
      r--;
    }
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      vector<int> A;
      int n, ans = 0;
      cin >> n;
      for (int i = 1; i <= n; i++) {
        cin >> a[i];
        if (a[i] % 2 != i % 2) {
          ans++;
        }
      }
      if (ans != 0) {
        cout << -1 << endl;
      } else {
        for (int u = n; u >= 3; u -= 2) {
          int k, l;
          for (int i = 1; i <= n; i++) {
            if (a[i] == u) {
              k = i;
              break;
            }
          }
          A.push_back(k);
          solve(k);
          for (int i = 1; i <= n; i++) {
            if (a[i] == u - 1) {
              l = i;
              break;
            }
          }
          A.push_back(l - 1);
          solve(l - 1);
          A.push_back(u);
          solve(u);
          A.push_back(u - l + 2);
          solve(u - l + 2);
          A.push_back(u);
          solve(u);
        }
        cout << A.size() << endl;
        for (auto v : A) {
          cout << v << " ";
        }
        cout << "" << endl;
      }
    }
  }
  