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.*;
public class Main {
    
    private void doit(){
        Scanner sc =new Scanner(System.in);
        while(sc.hasNext()){
            int n = sc.nextInt();
            if(n % 2 == 0){
                int [] data = new int[n+1];
                for(int i = 0; i <= n; i++){
                    data[i] = i;
                }
                
                for(int i = 1 ; i <= n; i+= 2){
                    int temp = data[i];
                    data[i] = data[i+1];
                    data[i+1] = temp;
                }
                
                System.out.print(data[1]);
                for(int i = 2; i <= n; i++){
                    System.out.print(" " + data[i]);
                }
                System.out.println();
            }
            else{
                System.out.println(-1);
            }
        }
    }

    public static void main(String[] args) {
        Main obj = new Main();
        obj.doit();
    }
}