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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
int countDigit(T n) {
  if (n < 0) n *= -1;
  return floor(log10(n) + 1);
}
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T lcm(T a, T b) {
  return a * b / gcd(a, b);
}
void fastscan(int &x) {
  bool neg = false;
  register int c;
  x = 0;
  c = getchar();
  if (c == '-') {
    neg = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) x = (x << 1) + (x << 3) + c - 48;
  if (neg) x *= -1;
}
int main() {
  int n;
  scanf("%d", &n);
  if (n & 1) {
    printf("-1");
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (i & 1)
      printf("%d ", i);
    else
      printf("%d ", i + 2);
  }
  return 0;
}