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.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.Scanner;

public class Perestanovka {

	public static int[] getArray(int size) {
		int[] array = new int[size];
		for (int i = 0; i < size; i++) {
			array[i] = sr.nextInt();
		}
		return array;
	}

	public static int[][] getMatrix(int sizeI, int sizeJ) {
		int[][] array = new int[sizeI][sizeJ];
		for (int i = 0; i < sizeI; i++) {
			for (int j = 0; j < sizeJ; j++) {
				array[i][j] = sr.nextInt();
			}
		}
		return array;
	}

	static Scanner sr = new Scanner(System.in);

	public static void main(String[] args) {

		int n = sr.nextInt();
		if(n != 1 && n%2 ==0){
			for (int i = 1; i <= n; i++) {
				System.out.print((n - i + 1) + " ");
			}
		}
		else{
			System.out.println(-1);
		}
	}
}