A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.

The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.

Input

The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.

Output

Print a single integer — the number of different possible locations after exactly n steps.

Examples

Input


1


Output


4


Input


2


Output


4


Input


3


Output


12

Note

In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.

In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.

from __future__ import division
import sys
input = sys.stdin.readline
import math
from math import sqrt, floor, ceil
from collections import Counter


############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
def insr2():
    s = input()
    return(s.split(" "))

def prime_factorization(n):

    if n == 1:
        return [1]

    ans=[]
    i = 2
    cap = sqrt(n)
    while i <= cap:
        if n % i == 0:
            ans.append(i)
            n = n//i
            cap=sqrt(n)
        else:
            i += 1
    if n > 1:
        ans.append(n)
    return ans

def binomial(n, k):
    if n == 1 or n == k:
        return 1

    if k > n:
        return 0       
    else:
        a = math.factorial(n)
        b = math.factorial(k)
        c = math.factorial(n-k)
        div = a // (b * c)
        return div  

"""
assuming that n is even, n%2==0:
we make n//2 moves horizontally and n//2 moves vertically
and these moves can be thought of as independent.

if n = 10, we make 5 moves left-right and 5 moves up-down.
when making 5 up-down moves, we can end up in 6 positions, we can choose 0up5down, ...., 5up0down.
same with left right, so we get 6*6

if n is odd e..g n = 11, we can get 6 updown and 5 leftright or 6 leftright and 5 updown.
we cannot end in the same positions in both approaches since parity.
so we have 6*7 times 2 there

e.g.
for n = 3:
    2 * (2) * (3) = 6

"""

n = inp()

if n % 2 == 0:
    ans = (n//2 + 1)**2
else:
    ans = 2 * ((n+1)//2) * ((n+1)//2 + 1)

print ans