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 
//package codeforces;

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.Closeable;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.ListIterator;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class A implements Closeable {
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    A() throws IOException {
//        reader = new BufferedReader(new FileReader("input.txt"));
//        writer = new PrintWriter(new FileWriter("output.txt"));
    }

    StringTokenizer stringTokenizer;

    String next() throws IOException {
        while (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
            stringTokenizer = new StringTokenizer(reader.readLine());
        }
        return stringTokenizer.nextToken();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    private int MOD = 1000 * 1000 * 1000 + 7;

    int sum(int a, int b) {
        a += b;
        return a >= MOD ? a - MOD : a;
    }

    int product(int a, int b) {
        return (int) (1l * a * b % MOD);
    }

    int pow(int x, int k) {
        int result = 1;
        while (k > 0) {
            if (k % 2 == 1) {
                result = product(result, x);
            }
            x = product(x, x);
            k /= 2;
        }
        return result;
    }

    int inv(int x) {
        return pow(x, MOD - 2);
    }

    void solve() throws IOException {
        int n = nextInt();
        if(n % 4 != 0 && n % 4 != 1) {
            writer.println(-1);
            return;
        }
        int[] p = new int[n];
        for(int i = 1; i < n / 2; i += 2) {
            int x = i, y = n - x + 1, u = i + 1, v = n - u + 1;
            //x -> u -> y -> v
            p[x - 1] = u;
            p[u - 1] = y;
            p[y - 1] = v;
            p[v - 1] = x;
        }
        if(n % 2 == 1) {
            p[n / 2] = n / 2 + 1;
        }
        for (int x : p) {
            writer.print(x + " ");
        }
    }

    public static void main(String[] args) throws IOException {
        try (A a = new A()) {
            a.solve();
        }
    }

    @Override
    public void close() throws IOException {
        reader.close();
        writer.close();
    }
}