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: 
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Scanner;
  
  public class E {
      public static void main(String[] args) {
          Scanner s = new Scanner(System.in);
          int lines = s.nextInt();
  
          s.nextLine();
  
          for (int i = 0; i < lines; i += 1) {
              solve(s.nextInt(), s);
          }
      }
  
  
      public static void solve(int n, Scanner s) {
          int a[] = new int[n];
          boolean stop = false;
          for (int i = 0; i < n; i++) {
              a[i] = s.nextInt();
              if (((a[i] - i) % 2 + 2) % 2 != 1) stop = true;
          }
  
          if (stop) {
              System.out.println(-1);
              return;
          }
  
          ArrayList<Integer> erg = new ArrayList<>();
          int l;
          int even = -1;
          int odd = -1;
          for (int len = n; len > 1; len -= 2) {
              for (int i = 0; i < len; i++) {
                  if (a[i] == len) odd = i + 1;
                  if (a[i] == len - 1) even = i + 1;
              }
  
  
              erg.add(odd);
              mirror(a, odd);
  
              if (even < odd) {
                  even = odd + 1 - even;
              }
  
              erg.add(even - 1);
              mirror(a, even - 1);
              odd = even - 1;
  
              erg.add(len);
              mirror(a, len);
  
              odd = len + 1 - odd;
  
              erg.add(odd);
              mirror(a, odd);
  
              erg.add(len);
              mirror(a, len);
          }
  
          boolean change = true;
          for (int i = erg.size() - 1; i >= 1; i--) {
              if (erg.get(i) == 1) {
                  erg.remove(i);
                  if (i > erg.size()) i = erg.size();
              }
              if (erg.get(i).equals(erg.get(i - 1))) {
                  erg.remove(i);
                  erg.remove(i - 1);
                  if (i > erg.size()) i = erg.size();
              }
          }
  
          System.out.println(erg.size());
          System.out.println(erg.toString().replace("[", "").replace("]", "").replace(",", ""));
      }
  
      public static void mirror(int a[], int n) {
          int m[] = new int[n];
          for (int i = 0; i < n; i++) {
              m[i] = a[n - i - 1];
          }
          for (int i = 0; i < n; i++) {
              a[i] = m[i];
          }
      }
  }
  