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.io.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.util.Arrays;
  import java.util.StringTokenizer;
  import java.io.IOException;
  import java.io.BufferedReader;
  import java.io.InputStreamReader;
  import java.io.InputStream;
  
  /**
   * Built using CHelper plug-in
   * Actual solution is at the top
   *
   * @author lucasr
   */
  public class Main {
  	public static void main(String[] args) {
  		InputStream inputStream = System.in;
  		OutputStream outputStream = System.out;
  		MyScanner in = new MyScanner(inputStream);
  		PrintWriter out = new PrintWriter(outputStream);
  		CBottomTierReversals solver = new CBottomTierReversals();
  		int testCount = Integer.parseInt(in.next());
  		for (int i = 1; i <= testCount; i++)
  			solver.solve(i, in, out);
  		out.close();
  	}
  
  	static class CBottomTierReversals {
  		public static MyScanner sc;
  		public static PrintWriter out;
  
  		public void solve(int testNumber, MyScanner sc, PrintWriter out) {
  			CBottomTierReversals.sc = sc;
  			CBottomTierReversals.out = out;
  			int n = sc.nextInt();
  			int[] perm = sc.nextIntArray(n);
  			for (int i = 0; i < n; i++) {
  				perm[i]--;
  			}
  			boolean ok = true;
  			for (int i = 0; i < n; i++) {
  				if (i % 2 != perm[i] % 2) ok = false;
  			}
  			if (!ok) {
  				out.println(-1);
  				return;
  			}
  			IntArray ans = new IntArray();
  
  			for (int i = n - 1; i > 0; i -= 2) {
  				toFirst(perm, ans, i);
  				firstToMatch(perm, ans);
  			}
  
  			out.println(ans.size());
  			for (int i = 0; i < ans.size(); i++) {
  				out.print(ans.get(i) + " ");
  			}
  			out.println();
  		}
  
  		static void rev(int[] p, IntArray op, int cant) {
  			if (cant > 1) {
  				op.add(cant);
  				ArrayUtils.reverse(p, 0, cant - 1);
  			}
  		}
  
  		static void toFirst(int[] perm, IntArray ops, int val) {
  			if (perm[0] == val) return;
  			for (int i = 0; i < perm.length; i++) {
  				if (perm[i] == val) {
  					rev(perm, ops, i + 1);
  					return;
  				}
  			}
  		}
  
  		static void firstToMatch(int[] perm, IntArray ops) {
  			for (int i = 1; i < perm.length; i += 2) {
  				if (perm[0] - 1 == perm[i]) {
  					int me = perm[0];
  					rev(perm, ops, i);
  					rev(perm, ops, i + 2);
  					rev(perm, ops, 3);
  					rev(perm, ops, me + 1);
  					return;
  				}
  			}
  		}
  
  	}
  
  	static class MyScanner {
  		private BufferedReader br;
  		private StringTokenizer tokenizer;
  
  		public MyScanner(InputStream is) {
  			br = new BufferedReader(new InputStreamReader(is));
  		}
  
  		public String next() {
  			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
  				try {
  					tokenizer = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					throw new RuntimeException(e);
  				}
  			}
  			return tokenizer.nextToken();
  		}
  
  		public int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		public int[] nextIntArray(int n) {
  			int[] ret = new int[n];
  			for (int i = 0; i < n; i++) {
  				ret[i] = nextInt();
  			}
  			return ret;
  		}
  
  	}
  
  	static class ArrayUtils {
  		public static void reverse(int[] array, int from, int to) {
  			while (from < to) {
  				int tmp = array[from];
  				array[from] = array[to];
  				array[to] = tmp;
  				from++;
  				to--;
  			}
  		}
  
  	}
  
  	static class IntArray {
  		int[] arr;
  		int size;
  
  		public IntArray() {
  			arr = new int[4];
  		}
  
  		public void add(int val) {
  			if (size == arr.length) {
  				arr = Arrays.copyOf(arr, 2 * arr.length);
  			}
  			arr[size++] = val;
  		}
  
  		public int get(int pos) {
  			return arr[pos];
  		}
  
  		public int size() {
  			return size;
  		}
  
  		public int[] toArray() {
  			return Arrays.copyOf(arr, size);
  		}
  
  		public String toString() {
  			return "IntArray " + Arrays.toString(toArray());
  		}
  
  	}
  }
  
  