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