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 

这道题就是找下规律就可以了，从n=3开始写几组出来，会发现奇数时是写不出来的，而当偶数是规律是从1开始每两个数相邻的数交换位置。。。。。。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
	int a[105],b[105];
	int n,i,j,k;
	j=k=0;
	scanf("%d",&n);
	if(n%2==1)
	{
		printf("-1");
	}
	else
	{
		for(i=1;i<=n;i++)
		{
			if(i%2==0)
			{
				a[k++]=i;
			}
			else
			{
				b[j++]=i;
			}
		}
		for(i=0;i<n/2;i++)
		{
			printf("%d %d ",a[i],b[i]);
		}

	}
	return 0;