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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.StringTokenizer;
  
  public class C {
  	
  	static ArrayList<Integer> answers=new ArrayList<>();
  
  	public static void main(String[] args) {
  		FastScanner fs=new FastScanner();
  		int T=fs.nextInt();
  		outer: for (int tt=0; tt<T; tt++) {
  			answers.clear();
  			int n=fs.nextInt();
  			int[] a=fs.readArray(n);
  			//TODO: check if possible;
  			for (int i=0; i<a.length; i++) {
  				if (a[i]%2==i%2) {
  					System.out.println("-1");
  					continue outer;
  				}
  			}
  			
  			if (n%2==0) throw null;
  			for (int biggerVal=n, smallerVal=n-1, gotoPos=n-1; 
  					biggerVal>1; 
  					biggerVal-=2, smallerVal-=2, gotoPos-=2) {
  				swap(indof(biggerVal, a), a);
  				swap(indof(smallerVal, a)-1, a);
  				swap(indof(smallerVal, a)+1, a);
  				swap(2, a);
  				swap(gotoPos, a);
  			}
  			for (int i=0; i<n; i++) if (a[i]!=i+1) throw null;
  			System.out.println(answers.size());
  			for (int i:answers)
  				System.out.print(i+" ");
  			System.out.println();
  		}
  	}
  	
  	static int indof(int x, int[] a) {
  //		System.out.println("Looking for "+x+" in "+Arrays.toString(a));
  		for (int i=0; i<a.length; i++) if (a[i]==x) return i;
  		throw null;
  	}
  	
  	static void swap(int prefixSize, int[] a) {
  		prefixSize++;
  		if (prefixSize%2!=1) throw null;
  		int[] newA=a.clone();
  		for (int i=0; i<prefixSize; i++) {
  			a[i]=newA[prefixSize-1-i];
  		}
  		answers.add(prefixSize);
  	}
  
  	static void sort(int[] a) {
  		ArrayList<Integer> l=new ArrayList<>();
  		for (int i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
  	
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  	}
  
  	
  }
  