#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  int a[n];
  if (n % 2 != 0)
    cout << -1;
  else {
    for (int i = 0; i < n; ++i) {
      a[i] = i + 1;
    }
    for (int i = 0; i < n - 1; i = i + 2) {
      cout << a[i + 1] << ' ' << a[i] << ' ';
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
int main() {
  int n;
  cin >> n;
  int a[n];
  if (n % 2 != 0)
    cout << -1;
  else {
    for (int i = 0; i < n; ++i) {
      a[i] = i + 1;
    }
    for (int i = 0; i < n - 1; i = i + 2) {
      cout << a[i + 1] << ' ' << a[i] << ' ';
    }
  }
  return 0;
}

```