#include <bits/stdc++.h>
int main() {
  int n, i;
  scanf("%d", &n);
  if (n % 2 != 0)
    printf("-1");
  else
    for (i = 2; i <= n; i = i + 2) printf("%d %d ", i, i - 1);
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
int main() {
  int n, i;
  scanf("%d", &n);
  if (n % 2 != 0)
    printf("-1");
  else
    for (i = 2; i <= n; i = i + 2) printf("%d %d ", i, i - 1);
}

```