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.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author @zhendeaini6001
 */
public class A {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        solver.solve(1, in, out);
        out.close();
	}
}

class TaskA {
    public void solve(int testNumber, Scanner in, PrintWriter out) {
    	int n = in.nextInt();
    	if (n % 2 == 1){
    		out.println("-1");
    	}else{
    		for (int i = 1; i <= n; ++i){
    			if (i % 2 == 0){
    				out.print(i-1);
    				if (i != n){
    					out.print(" ");
    				}
    			}else{
    				out.print(i+1);
    				out.print(" ");
    			}
    		}
    	}
    	out.println();
    	return;
    }
}