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.util.Scanner;


public class Main {


	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n=scan.nextInt();
		int arr[]=new int[n];
		for (int i = 0; i < arr.length; i++) {
			arr[i]=i+1;
		}
		if(n==1)System.out.println("-1");
		else{
		if(n%2==0){
			for (int i = 0; i < n-1; i++) {
				int temp=arr[i];
				arr[i]=arr[i+1];
				arr[i+1]=temp;
				i++;
			}
			for (int i = 0; i < arr.length; i++) {
				System.out.print(arr[i]+" ");
			}
		}else{
			System.out.println("-1");
			/*for (int i = 0; i < n-3; i++) {
				int temp=arr[i];
				arr[i]=arr[i+1];
				arr[i+1]=temp;
				i++;
			}
			int t =arr[arr.length-1];
			arr[arr.length-1]=arr[arr.length-3];
			arr[arr.length-3]=t;
			t=arr[arr.length-2];
			arr[arr.length-2]=arr[arr.length-3];
			arr[arr.length-3]=t;
			for (int i = 0; i < arr.length; i++) {
				System.out.print(arr[i]+" ");
			}*/
		}
			
		}
		
	}

}