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.*;
  import java.io.*;
  
  public class _1561_E {
      public static void main(String[] args) throws IOException {
          Reader in = new Reader();
          PrintWriter out = new PrintWriter(System.out);
          int t = in.nextInt();
          while(t-- > 0) {
              int n = in.nextInt();
              int[] a = new int[n];
              boolean possible = true;
              for(int i = 0; i < n; i++) {
                  a[i] = in.nextInt() - 1;
                  if(a[i] % 2 != i % 2) {
                      possible = false;
                  }
              }
              if(!possible) {
                  out.println(-1);
              }else {
                  int res = 0;
                  StringBuilder sb = new StringBuilder();
                  for(int i = n - 1; i > 0; i -= 2) {
  
                      int ind = findVal(a, i);
                      reverse(a, ind);
                      sb.append(ind + 1);
                      sb.append(' ');
  
                      ind = findVal(a, i - 1);
                      reverse(a, ind - 1);
                      sb.append(ind);
                      sb.append(' ');
  
                      reverse(a, ind + 1);
                      sb.append(ind + 2);
                      sb.append(' ');
  
                      reverse(a, 2);
                      sb.append(3);
                      sb.append(' ');
  
                      reverse(a, i);
                      sb.append(i + 1);
                      sb.append(' ');
  
                      res += 5;
  
                  }
                  for(int i = 0; i < n; i++) {
                      if(i != a[i]) {
                          possible = false;
                          break;
                      }
                  }
                  if(!possible) {
                      out.println(-1);
                  }else {
                      out.println(res);
                      out.println(sb.toString());
                  }
              }
          }
          in.close();
          out.close();
      }
      static void reverse(int[] a, int ind) {
          for(int i = 0; i <= ind / 2; i++) {
              int temp = a[i];
              a[i] = a[ind - i];
              a[ind - i] = temp;
          }
      }
      static int findVal(int[] a, int val) {
          for(int i = 0; i < a.length; i++) {
              if(a[i] == val) {
                  return i;
              }
          }
          return -1;
      }
      static class Reader {
          final private int BUFFER_SIZE = 1 << 16;
          private DataInputStream din;
          private byte[] buffer;
          private int bufferPointer, bytesRead;
  
          public Reader() {
              din = new DataInputStream(System.in);
              buffer = new byte[BUFFER_SIZE];
              bufferPointer = bytesRead = 0;
          }
  
          public Reader(String file_name) throws IOException {
              din = new DataInputStream(new FileInputStream(file_name));
              buffer = new byte[BUFFER_SIZE];
              bufferPointer = bytesRead = 0;
          }
  
          public String readLine() throws IOException {
              byte[] buf = new byte[64]; // line length
              int cnt = 0, c;
              while ((c = read()) != -1) {
                  if (c == '\n') {
                      if (cnt != 0) {
                          break;
                      } else {
                          continue;
                      }
                  }
                  buf[cnt++] = (byte) c;
              }
              return new String(buf, 0, cnt);
          }
  
          public int nextInt() throws IOException {
              int ret = 0;
              byte c = read();
              while (c <= ' ') {
                  c = read();
              }
              boolean neg = (c == '-');
              if (neg) {
                  c = read();
              }
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
              if (neg) {
                  return -ret;
              }
              return ret;
          }
  
          public long nextLong() throws IOException {
              long ret = 0;
              byte c = read();
              while (c <= ' ') {
                  c = read();
              }
              boolean neg = (c == '-');
              if (neg) {
                  c = read();
              }
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
              if (neg) {
                  return -ret;
              }
              return ret;
          }
  
          public double nextDouble() throws IOException {
              double ret = 0, div = 1;
              byte c = read();
              while (c <= ' ') {
                  c = read();
              }
              boolean neg = (c == '-');
              if (neg) {
                  c = read();
              }
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
              if (c == '.') {
                  while ((c = read()) >= '0' && c <= '9') {
                      ret += (c - '0') / (div *= 10);
                  }
              }
              if (neg) {
                  return -ret;
              }
              return ret;
          }
  
          private void fillBuffer() throws IOException {
              bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
              if (bytesRead == -1) {
                  buffer[0] = -1;
              }
          }
  
          private byte read() throws IOException {
              if (bufferPointer == bytesRead) {
                  fillBuffer();
              }
              return buffer[bufferPointer++];
          }
  
          public void close() throws IOException {
              if (din == null) {
                  return;
              }
              din.close();
          }
      }
  }
  