Problem: 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.

Solution (in PYTHON3):

Solution:
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**7)
# sys.stdin = open('inpy.txt', 'r')
# sys.stdout = open('outpy.txt', 'w')
from sys import stdin, stdout
import bisect  # c++ upperbound
import math
import heapq
i_m = 9223372036854775807


def inin():
    return int(input())


def stin():
    return input()


def spin():
    return map(int, stin().split())


def lin():  # takes array as input
    return list(map(int, stin().split()))


def matrix(n):
    #matrix input
    return [list(map(int, input().split()))for i in range(n)]

################################################


def string2intlist(s):
    return list(map(int, s))


def calculate_sum(a, N):  # sum of a to N
    # Number of multiples
    m = N / a
    # sum of first m natural numbers
    sum = m * (m + 1) / 2
    # sum of multiples
    ans = a * sum
    return ans


def series(N):
    return (N*(N+1))//2


def count2Dmatrix(i, list):
    return sum(c.count(i) for c in list)


def modinv(n, p):
    return pow(n, p - 2, p)


def nCr(n, r):
    i = 1
    while i < r:
        n *= (n - i)
        i += 1
    return n // math.factorial(r)


def LCM(x, y):
    return (x * y) // math.gcd(x, y)


def Divisors(n):
    l = []
    for i in range(1, int(math.sqrt(n) + 1)):
        if (n % i == 0):
            if (n // i == i):
                l.append(i)
            else:
                l.append(i)
                l.append(n//i)
    return l


def isprime(n):
    if n == 2 or n == 3:
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False
    for i in range(5, int(math.sqrt(n))+1, 6):
        if n % i == 0 or n % (i+2) == 0:
            return False
    return True


prime = []


def SieveOfEratosthenes(n):
    global prime
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    f = []
    for p in range(2, n):
        if prime[p]:
            f.append(p)
    return f


q = []


def dfs(n, d, v, c):
    global q
    v[n] = 1
    x = d[n]
    q.append(n)
    j = c
    for i in x:
        if i not in v:
            f = dfs(i, d, v, c+1)
            j = max(j, f)
            # print(f)
    return j
# d = {}


def knapSack(W, wt, val, n):
    K = [[0 for x in range(W + 1)] for x in range(n + 1)]

    # Build table K[][] in bottom up manner
    for i in range(n + 1):
        for w in range(W + 1):
            if i == 0 or w == 0:
                K[i][w] = 0
            elif wt[i-1] <= w:
                K[i][w] = max(val[i-1] + K[i-1][w-wt[i-1]],  K[i-1][w])
            else:
                K[i][w] = K[i-1][w]

    return K[n][W]


def modularExponentiation(x, n):
    M = 10**9+7
    if(n == 0):
        return 1
    elif (n % 2 == 0):  # n is even
        return modularExponentiation((x*x) % M, n//2)
    else:  # n is odd
        return (x * modularExponentiation((x * x) % M, (n - 1) // 2)) % M


def modInverse(a, m):
    m0 = m
    y = 0
    x = 1

    if (m == 1):
        return 0

    while (a > 1):

        # q is quotient
        q = a // m

        t = m

        # m is remainder now, process
        # same as Euclid's algo
        m = a % m
        a = t
        t = y

        # Update x and y
        y = x - q * y
        x = t

    # Make x positive
    if (x < 0):
        x = x + m0

    return x


"""*******************************************************"""
# for _ in range(inin()):
n = inin()

if n % 2 == 0:
    print((n // 2 + 1)** 2)
else:
    ans = 0
    for i in range(n + 1, 0, -2):
        ans += 2 * i
    
    print(ans)