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 int[] per;

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		per = new int[n];
		if (n > 1) {
			for (int i = 1; i <= n; i++) {
				for (int j = 1; j <= n; j++) {
					if (i != j && check(j)) {
						per[i - 1] = j;
						break;
					}
				}
			}
			if (!check(0)) {
				System.out.println(-1);
				System.exit(0);
			}
			for (int i = 0; i < per.length; i++) {
				System.out.print(per[i] + " ");
			}
		} else {
			System.out.println(-1);
		}
	}

	public static boolean check(int value) {
		for (int i = 0; i < per.length; i++) {
			if (per[i] == value) {
				return false;
			}

		}
		return true;
	}

}