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 
#include <bits/stdc++.h>
using namespace std;
long long fact(long long n) {
  long long f = 1;
  for (long long i = n; i >= 1; i--) f *= i;
  return f;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int divisor(int n) {
  int sum = 0;
  if (n / sqrt(n) == sqrt(n)) sum--;
  for (int i = 1; i <= sqrt(n); i++)
    if (n % i == 0) sum += 2;
  return sum;
}
bool prime(long long n) {
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
bool palindrome_str(string n) {
  int x = 0;
  int m = n.size() / 2;
  int r = n.size();
  for (int i = 0; i < m; i++)
    if (n[i] == n[r - i - 1]) x++;
  if (x == m) return true;
  return false;
}
bool palindrome_num(int n) {
  int x = 0, a[10000], r = 0;
  while (n) {
    a[r++] = n % 10;
    n /= 10;
  }
  for (int i = 0; i < r / 2; i++)
    if (a[i] == a[r - i - 1]) x++;
  if (x == r / 2) return true;
  return false;
}
void fast() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  fast();
  int r;
  cin >> r;
  if (r % 2 != 0) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i < r; i += 2) cout << i + 1 << " " << i << " ";
  cin >> r;
}
