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.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.Set;

public class Main {
	public static void main(String[] args) {
		Task task = new Task();
		task.getData();
		task.solve();
		task.printAnswer();
	}
}

class Task {
	int n;

	String answer = "";

	public void getData() {
		Scanner console = new Scanner(System.in);

		n = console.nextInt();

		console.close();
	}

	public void solve() {
		if (n % 2 != 0) {
			answer = "-1";
			return;
		}
		for (int i = 1; i <= n; ++i) {
			answer += n + 1 - i + " ";
		}
	}

	public void printAnswer() {
		System.out.print(answer);
	}
}

class Point {
	int f;
	int t;

	Point(int f, int t) {
		this.f = f;
		this.t = t;
	}
}