#include <bits/stdc++.h>
using namespace std;
bool prime[1001];
void tk_rwing() {
  memset(prime, true, sizeof(prime));
  for (long long i = 2; i * i <= 1000; i++) {
    if (prime[i]) {
      for (long long j = i * i; j <= 1000; j += i) {
        prime[j] = false;
      }
    }
  }
}
long long prime_c[1001];
void tk_cout() {
  long long c = 0;
  for (long long i = 1; i < 1001; i++) {
    if (prime[i]) c++;
    prime_c[i] = c;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  if (n % 2 == 1)
    cout << -1;
  else {
    long long c = 0;
    for (long long i = 0; i < n / 2; i++) {
      cout << 2 + c << ' ' << 1 + c << ' ';
      c += 2;
    }
  }
}

### 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;
bool prime[1001];
void tk_rwing() {
  memset(prime, true, sizeof(prime));
  for (long long i = 2; i * i <= 1000; i++) {
    if (prime[i]) {
      for (long long j = i * i; j <= 1000; j += i) {
        prime[j] = false;
      }
    }
  }
}
long long prime_c[1001];
void tk_cout() {
  long long c = 0;
  for (long long i = 1; i < 1001; i++) {
    if (prime[i]) c++;
    prime_c[i] = c;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  if (n % 2 == 1)
    cout << -1;
  else {
    long long c = 0;
    for (long long i = 0; i < n / 2; i++) {
      cout << 2 + c << ' ' << 1 + c << ' ';
      c += 2;
    }
  }
}

```