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.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class A {

    static BufferedReader in;
    static StringTokenizer st;

    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        int n = nextInt();
        if (n % 2 == 1) {
            System.out.println("-1");
            return;
        }
        for (int i = 1; i <= n; i++) {
            if (i % 2 == 0) {
                System.out.print((i - 1) + " ");
            } else {
                System.out.print((i + 1) + " ");
            }
        }
    }

}