#include <bits/stdc++.h>
using namespace std;
const long double PI = 4 * atan((long double)1);
const long long INF = 1e18;
const long long mod = 1e9 + 7;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tt = 1;
  for (long long qq = 1; qq <= (tt); ++qq) {
    long long n;
    cin >> n;
    if (n & 1) {
      cout << -1;
    } else {
      long long i = 2;
      while (i <= n) {
        cout << i << " " << i - 1 << " ";
        i += 2;
      }
    }
  }
}

### Prompt

Your task is to create a cpp solution to the following 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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long double PI = 4 * atan((long double)1);
const long long INF = 1e18;
const long long mod = 1e9 + 7;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tt = 1;
  for (long long qq = 1; qq <= (tt); ++qq) {
    long long n;
    cin >> n;
    if (n & 1) {
      cout << -1;
    } else {
      long long i = 2;
      while (i <= n) {
        cout << i << " " << i - 1 << " ";
        i += 2;
      }
    }
  }
}

```