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 Solution {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		int p = sc.nextInt();

		if (p % 2 == 0){
			int[] arr = new int[p];
			permutation(arr);
		} else {
			System.out.println(-1);
		}
	}

	public static void permutation(int[] arr){

		for (int i = 0; i < arr.length; i++) {
			arr[i] = i+1;
		}

		for (int i = 1; i < arr.length; i = i + 2) {
			int tmp = arr[i];
			arr[i] = arr[i-1];
			arr[i-1] = tmp;
		}

		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < arr.length; i++) {
			sb.append(arr[i]).append(" ");
		}

		System.out.println(sb);
	}

}