A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).

A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.

You have integer n. Find some lucky permutation p of size n.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.

Output

Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.

Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.

If there are multiple answers, you can print any of them.

Examples

Input

1


Output

1 


Input

2


Output

-1


Input

4


Output

2 4 1 3 


Input

5


Output

2 5 3 1 4 
import java.io.InputStreamReader;
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.io.BufferedWriter;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.Writer;
import java.util.StringTokenizer;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Lokesh Khandelwal
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		OutputWriter out = new OutputWriter(outputStream);
		TaskA solver = new TaskA();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskA {
    public void solve(int testNumber, InputReader in, OutputWriter out) {
        int n=in.nextInt();
        int i,a[]=new int[n+1];
        if(n==1)
        {
            out.printLine(1);
            return;
        }
        if(n%4==2||n%4==3)
        {
            out.printLine(-1);
            return;
        }

        int times=n/4;
        int c=1;
        while (c<=times)
        {
            a[2*c-1]=2*c;
            a[2*c]=n-(2*c-1)+1;
            a[n-(2*c-1)+1]=n-2*c+1;
            a[n-2*c+1]=n-(n-2*c+2)+1;
            c++;
        }
        if(n%4==1)
            a[n/2+1]=n/2+1;
        for(i=1;i<n;i++)
        {
            out.print(a[i]+" ");
        }
        out.printLine(a[i]);
    }
}

class InputReader
{
    BufferedReader in;
    StringTokenizer tokenizer=null;

    public InputReader(InputStream inputStream)
    {
        in=new BufferedReader(new InputStreamReader(inputStream));
    }
    public String next()
    {
        try{
            while (tokenizer==null||!tokenizer.hasMoreTokens())
            {
                tokenizer=new StringTokenizer(in.readLine());
            }
            return tokenizer.nextToken();
        }
        catch (IOException e)
        {
            return null;
        }
    }
    public int nextInt()
    {
        return Integer.parseInt(next());
    }
    }

class OutputWriter {
	private final PrintWriter writer;

	public OutputWriter(OutputStream outputStream) {
		writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
	}

	public OutputWriter(Writer writer) {
		this.writer = new PrintWriter(writer);
	}

	public void print(Object...objects) {
		for (int i = 0; i < objects.length; i++) {
			if (i != 0)
				writer.print(' ');
			writer.print(objects[i]);
		}
	}

	public void printLine(Object...objects) {
		print(objects);
		writer.println();
	}

	public void close() {
		writer.close();
	}
}

