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 optimize("Ofast")
  using namespace std;
  void printn(vector<int> nums) {
    for (int a : nums) {
      cout << a << " ";
    }
    cout << '\n';
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t, n;
    cin >> t;
    for (int test = 0; test < t; test++) {
      cin >> n;
      vector<int> nums(n);
      bool Pos = true;
      for (int a = 0; a < n; a++) {
        cin >> nums[a];
        if ((nums[a] + 1 - a) % 2 != 0) {
          Pos = false;
        }
      }
      if (!Pos) {
        cout << -1 << '\n';
        continue;
      }
      int pindex;
      int oindex;
      int act = n;
      cout << 5 * (n - 1) / 2 << '\n';
      while (act > 1) {
        for (int a = 0; a < n; a++) {
          if (nums[a] == act) {
            oindex = a;
          }
          if (nums[a] == act - 1) {
            pindex = a;
          }
        }
        cout << oindex + 1 << " ";
        reverse(nums.begin(), nums.begin() + oindex + 1);
        if (oindex > pindex) {
          pindex = oindex - pindex;
        }
        oindex = 0;
        cout << pindex << " ";
        reverse(nums.begin(), nums.begin() + pindex);
        cout << pindex + 2 << " ";
        reverse(nums.begin(), nums.begin() + pindex + 2);
        cout << 3 << " ";
        reverse(nums.begin(), nums.begin() + 3);
        cout << act << " ";
        reverse(nums.begin(), nums.begin() + act);
        act -= 2;
      }
      cout << '\n';
    }
    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
  }
  