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;
  class Solution {
    int n;
    vector<int> a, pos, res;
    void reverse(int k) {
      if (k == 1) return;
      assert(k % 2 == 1);
      vector<int> tmp(a.begin(), a.begin() + k + 1);
      for (int i = 1; i < k + 1; i++) {
        a[i] = tmp[k - i + 1];
        pos[a[i]] = i;
      }
      res.push_back(k);
    }
  
   public:
    Solution(vector<int> &v) { a = v; }
    pair<bool, vector<int> > Solve() {
      n = a.size() - 1;
      for (int i = 1; i < n + 1; i++)
        if ((a[i] + i) % 2 != 0) {
          return make_pair(false, vector<int>());
        }
      res.clear();
      pos.assign(n + 1, 0);
      for (int i = 1; i < n + 1; i++) pos[a[i]] = i;
      for (int i = n; i > 1; i -= 2) {
        int j = i - 1;
        if (pos[i] == i && pos[j] == j) continue;
        if (pos[i] > pos[j]) {
          reverse(pos[i]);
          assert(pos[i] < pos[j]);
        }
        if (pos[i] != pos[j] - 1) {
          reverse(pos[i]);
          reverse(pos[j] - 1);
          assert(pos[i] == pos[j] - 1);
        }
        if (pos[i] != 1 && pos[j] != 2) {
          reverse(pos[j] + 1);
          reverse(3);
          assert(pos[i] == 1);
          assert(pos[j] == 2);
        }
        reverse(i);
        assert(pos[i] == i);
        assert(pos[j] == j);
      }
      if (pos[1] != 1) {
        cout << "failed: ";
        for (int i = 1; i < a.size(); i++) cout << a[i] << ' ';
        cout << "\n";
      }
      return make_pair(true, res);
    }
  };
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      vector<int> a(n + 1);
      for (int i = 1; i < n + 1; i++) cin >> a[i];
      Solution obj(a);
      auto res = obj.Solve();
      if (res.first == false) {
        cout << "-1\n";
        continue;
      }
      cout << res.second.size() << '\n';
      for (int i = 0; i < res.second.size(); i++)
        cout << res.second[i] << " \n"[i == res.second.size() - 1];
    }
    return 0;
  }
  