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 PYTHON3):

Solution:
from math import ceil, log, floor, sqrt
import math	
	
	
k = 1	
def mod_expo(n, p, m):
	"""find (n^p)%m"""
	result = 1
	while p != 0:
		if p%2 == 1:
			result = (result * n)%m
		p //= 2
		n = (n * n)%m
	return result
	
def find_order(n):
	
	if n%2 == 0:
		#res = x for x in range(n, 0, -1)
		print(*[x for x in range(n, 0, -1)], sep=' ')
	else:
		print(-1)
	
	
	
t = 1
#t = int(input())
while t:
	t = t - 1
	k, g = 0, 0
	points = []
	n = int(input()) 
	#a = input()
	#b = input()
	#n, p, q, r = map(int, input().split()) 
	#n, m  = map(int, input().split())  
	#print(discover())
	# = map(int, input().split())
	#a = list(map(int, input().strip().split()))[:2*n]
	#w = list(map(int, input().strip().split()))[:k]
	#for i in range(3):
	#	x, y = map(int, input().split()) 
	#	points.append((x, y))
#	s = input()
	#if possible_phone_number(n, a):
	#	print("YES")
	#else:
	#	print("NO")
	find_order(n)
	#print(find_mx_teams(n, m))