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.*;
import java.util.*;

public class A {
	public static void main(String[] args) throws Exception {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		Scanner s = new Scanner(System.in);
		
		int n = Integer.parseInt(f.readLine());
		if(n == 1) {
			System.out.println(-1);
			return;
		}
		int[] ans = new int[n];
		int[] idx = new int[n+1];
		boolean[] used = new boolean[n];
		for(int i = 0; i < n; i++) {
			for(int j = 0; j < n; j++) {
				if(i == j) continue;
				if(!used[j]) {
					used[j] = true;
					ans[j] = i+1;
					idx[i+1] = j;
					break;
				}
			}
		}
		for(int i = 0; i < n; i++) {
			if(ans[i] == 0) {
				System.out.println(-1);
				return;
			}
		}
		for(int i = 0; i < n; i++) {
			if(i != n-1) out.print(ans[i] + " ");
			else out.println(ans[i]);
		}
		out.close();
	}
}