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 Main {
      public static void main(String args[]) {new Main().run();}
  
      FastReader in = new FastReader();
      PrintWriter out = new PrintWriter(System.out);
      void run() {
          for(int q=ni();q>0;q--){
              work();
          }
          out.flush();
      }
      long mod=1000000007;
      long gcd(long a,long b) {
          return a==0?b:gcd(b%a,a);
      }
      long inf=Long.MAX_VALUE/3;
      int n;
      int[] A;
      void work() {
          n=ni();
          A=new int[n+1];
          for(int i=1;i<=n;i++){
              A[i]=ni();
          }
          for(int i=1;i<=n;i++){
              if(i%2!=A[i]%2){
                  out.println(-1);
                  return;
              }
          }
          List<Integer> ret=new ArrayList<>();
          for(int i=n;i>1;i-=2){
              int idx=getIdx(i);
              handle(idx,ret);
              idx=getIdx(i-1);
              handle(idx-1,ret);
              handle(idx+1,ret);
              handle(3,ret);
              handle(i,ret);
          }
          out.println(ret.size());
          for(int r:ret){
              out.print(r+" ");
          }
          out.println();
      }
  
      private void handle(int idx, List<Integer> ret) {
          reverse(idx);
          ret.add(idx);
      }
  
      private void reverse(int idx) {
          for(int i=1,j=idx;i<j;i++,j--){
              int v=A[i];
              A[i]=A[j];
              A[j]=v;
          }
      }
  
      private int getIdx(int v) {
          for(int i=1;i<=n;i++){
              if(A[i]==v){
                  return i;
              }
          }
          return -2;
      }
  
      @SuppressWarnings("unused")
      private ArrayList<Integer>[] ng(int n, int m) {
          ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
          for(int i=0;i<n;i++) {
              graph[i]=new ArrayList<>();
          }
          for(int i=1;i<=m;i++) {
              int s=in.nextInt()-1,e=in.nextInt()-1;
              graph[s].add(e);
              graph[e].add(s);
          }
          return graph;
      }
  
      private ArrayList<long[]>[] ngw(int n, int m) {
          ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
          for(int i=0;i<n;i++) {
              graph[i]=new ArrayList<>();
          }
          for(int i=1;i<=m;i++) {
              long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
              graph[(int)s].add(new long[] {e,w});
              graph[(int)e].add(new long[] {s,w});
          }
          return graph;
      }
  
      private int ni() {
          return in.nextInt();
      }
  
      private long nl() {
          return in.nextLong();
      }
      private double nd() {
          return in.nextDouble();
      }
      private String ns() {
          return in.next();
      }
  
      private long[] na(int n) {
          long[] A=new long[n];
          for(int i=0;i<n;i++) {
              A[i]=in.nextLong();
          }
          return A;
      }
  
      private int[] nia(int n) {
          int[] A=new int[n];
          for(int i=0;i<n;i++) {
              A[i]=in.nextInt();
          }
          return A;
      }
  }
  
  class FastReader
  {
      BufferedReader br;
      StringTokenizer st;
      InputStreamReader input;//no buffer
      public FastReader()
      {
          br=new BufferedReader(new InputStreamReader(System.in));
      }
  
      public FastReader(boolean isBuffer)
      {
          if(!isBuffer){
              input=new InputStreamReader(System.in);
          }else{
              br=new BufferedReader(new InputStreamReader(System.in));
          }
      }
  
      public boolean hasNext(){
          try{
              String s=br.readLine();
              if(s==null){
                  return  false;
              }
              st=new StringTokenizer(s);
          }catch(IOException e){
              e.printStackTrace();
          }
          return true;
      }
  
      public String next()
      {
          if(input!=null){
              try {
                  StringBuilder sb=new StringBuilder();
                  int ch=input.read();
                  while(ch=='\n'||ch=='\r'||ch==32){
                      ch=input.read();
                  }
                  while(ch!='\n'&&ch!='\r'&&ch!=32){
                      sb.append((char)ch);
                      ch=input.read();
                  }
                  return sb.toString();
              }catch (Exception e){
                  e.printStackTrace();
              }
          }
          while(st==null || !st.hasMoreElements())//回车，空行情况
          {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
  
      public int nextInt()
      {
          return (int)nextLong();
      }
  
      public long nextLong() {
          try {
              if(input!=null){
                  long ret=0;
                  int b=input.read();
                  while(b<'0'||b>'9'){
                      b=input.read();
                  }
                  while(b>='0'&&b<='9'){
                      ret=ret*10+(b-'0');
                      b=input.read();
                  }
                  return ret;
              }
          }catch (Exception e){
              e.printStackTrace();
          }
          return Long.parseLong(next());
      }
  
      public double nextDouble()
      {
          return Double.parseDouble(next());
      }
  }