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.*;
import java.util.*;
public class Main
{
	public static void main(String[] args)throws IOException
	{
		new Main().start();
	}
	public void start()throws IOException
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;
		//int test=Integer.parseInt(br.readLine());
		//int len=Integer.parseInt(br.readLine());
		//int x = st.nextToken();
		//st = new StringTokenizer(br.readLine());
		//int mod = (int)(Math.pow(10,9))+7;
		//int count = 0;
		//3^n + 3* ((-1)^n)
		//System.out.println(mod);
		/*while(test --> 0){
			
		}*/
		int n = Integer.parseInt(br.readLine());
		if(n%2 == 1){
			System.out.println(-1);
			return;
		}
		int i=1;
		String ans = "";
		while(i<n){
			ans += (i+1+" "+i+" ");
			i+=2;
		}
		ans = ans.trim();
		System.out.print(ans);
	}
}