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() {
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      int a[2030];
      for (int i = 0; i < n; i++) {
        cin >> a[i];
        a[i]--;
      }
      bool f = true;
      for (int i = 0; i < n; i++) {
        if ((a[i] + i) % 2) f = false;
      }
      if (!f) {
        cout << -1 << endl;
        continue;
      }
      cout << n / 2 * 5 << endl;
      for (int i = n; i >= 3; i -= 2) {
        for (int j = 0; j < i; j++) {
          if (a[j] == i - 1) {
            cout << j + 1 << " ";
            reverse(a, a + j + 1);
            break;
          }
        }
        for (int j = 0; j < i; j++) {
          if (a[j] == i - 2) {
            cout << j << " ";
            reverse(a, a + j);
            cout << j + 2 << " ";
            reverse(a, a + j + 2);
            cout << 3 << " ";
            swap(a[0], a[2]);
            cout << i << " ";
            reverse(a, a + i);
            break;
          }
        }
      }
      cout << endl;
    }
  }
  