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 CPP):

Solution:
#include <bits/stdc++.h>
long long gcd(long long a, long long b) {
  if (a == 0 || b == 0) return (a + b);
  if (a > b) return (gcd(a % b, b));
  return (gcd(a, b % a));
}
long long lcm(long long a, long long b) { return ((a * b / gcd(a, b))); }
int sqr(int x) { return (x * x); }
void out() {
  puts("NO");
  exit(0);
}
using namespace std;
int n;
int p[100008];
int main() {
  cin >> n;
  for (int i = (int)(1); i < (int)(n + 1); ++i) p[i] = i;
  if (n % 2 == 1) {
    cout << "-1\n";
    return 0;
  }
  for (int i = (int)(1); i < (int)(n + 1); ++i) {
    swap(p[i], p[i + 1]);
    i++;
  }
  for (int i = (int)(n); i >= (int)(1); --i) cout << i << " ";
  cout << endl;
  return 0;
}