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.io.BufferedReader;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Collections;


public class PlayGround {

	public static void main(String[] args) throws Exception {
		Eduardo();

	}
	
	public static void Eduardo() throws IOException{

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int permutation = Integer.parseInt(br.readLine());
		if(permutation % 2 == 1){
			System.out.println(-1);
		}
		else{
			int[] array = new int[permutation];
			int count = 1;
			for(int i = 0; i < permutation; i++){
				if(count % 2 == 1){
					array[i+1] = count;

				}
				else{
					array[i-1] = count;
				}
				count++;
			}

			
			StringBuilder sb = new StringBuilder();
			for(int integer : array){
				sb.append(integer + " ");
			}
			System.out.println(sb);
			
		}

	}
}