#include <bits/stdc++.h>
using namespace std;
long long Bin_expo(long long n, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * n);
    }
    n = (n * n);
    b /= 2;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  if (n & 1)
    cout << -1 << "\n";
  else {
    for (long long i = 0; i < n; i++) {
      if (i % 2) {
        cout << i << " ";
      } else
        cout << i + 2 << " ";
    }
    cout << "\n";
  }
}

### 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>
using namespace std;
long long Bin_expo(long long n, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * n);
    }
    n = (n * n);
    b /= 2;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  if (n & 1)
    cout << -1 << "\n";
  else {
    for (long long i = 0; i < n; i++) {
      if (i % 2) {
        cout << i << " ";
      } else
        cout << i + 2 << " ";
    }
    cout << "\n";
  }
}

```