Problem: A permutation is a sequence of integers p1, p2, ..., pn, consisting of n distinct positive integers, each of them doesn't exceed n. Let's denote the i-th element of permutation p as pi. We'll call number n the size of permutation p1, p2, ..., pn.

Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation p that for any i (1 ≤ i ≤ n) (n is the permutation size) the following equations hold ppi = i and pi ≠ i. Nickolas asks you to print any perfect permutation of size n for the given n.

Input

A single line contains a single integer n (1 ≤ n ≤ 100) — the permutation size.

Output

If a perfect permutation of size n doesn't exist, print a single integer -1. Otherwise print n distinct integers from 1 to n, p1, p2, ..., pn — permutation p, that is perfect. Separate printed numbers by whitespaces.

Examples

Input

1


Output

-1


Input

2


Output

2 1 


Input

4


Output

2 1 4 3 

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.Vector;

public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		Scanner in = new Main().new Scanner(inputStream);
		Main solver = new Main();
		solver.solve2(in);
	}

	public void solve2(Scanner in) {
		int n = in.ni();
		if(n % 2 == 1) {
			System.out.println(-1);
		} else {
			for (int i = 1; i <= n / 2; i++) {
				System.out.print((i * 2) + " " + (i*2-1) + " ");
			}
		}
    } 
  
	class Scanner {
		BufferedReader br;
		StringTokenizer in;

		public Scanner(InputStream inputStream) {
			br = new BufferedReader(new InputStreamReader(inputStream));
		}

		public String next() {
			return hasMoreTokens() ? in.nextToken() : null;
		}

		public String readLine() {
			try {
				return br.readLine();
			} catch (Exception e) {
				e.printStackTrace();
				return null;
			}
		}

		boolean hasMoreTokens() {
			while (in == null || !in.hasMoreTokens()) {
				String s = readLine();
				if (s == null)
					return false;
				in = new StringTokenizer(s);
			}
			return true;
		}

		public String nextString() {
			return next();
		}

		public int ni() {
			return Integer.parseInt(nextString());
		}

		public double nd() {
			return Double.parseDouble(nextString());
		}

		public long nl() {
			return Long.parseLong(nextString());
		}

		public int[] nia(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = ni();
			}
			return a;
		}

		public long[] nla(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++) {
				a[i] = ni();
			}
			return a;
		}
	}

}