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 CPP):

Solution:
#pragma GCC optimize("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
#include <stack>
#include <deque>
#include <iterator>
#include <tuple>
#include <iomanip>
#include <valarray>
#include <limits>
#include <sstream>
#include <cassert>
#include <functional>
#include <numeric>
#define in insert
#define pb push_back
#define mp make_pair
#define int long long
#define fi first
#define double long double
#define se second
#define endl '\n'
#define pii pair<int,int>
#define pdd pair<double,double>
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);
#define mod 1000000007
#define MAXN  3000001
#define ll unsigned long long
using namespace std;
/*using namespace __gnu_pbds;
 typedef tree<int, null_type, less<pii>, rb_tree_tag,
 tree_order_statistics_node_update>
 ordered_set;*/
int spf[MAXN];
int fac[MAXN];
void sieve()//sieve
{
    spf[1] = 1;
    for (int i=2; i<MAXN; i++)
        spf[i] = i;
    for (int i=4; i<MAXN; i+=2)
        spf[i] = 2;
    for (int i=3; i*i<MAXN; i++)
    {
        if (spf[i]==i)
        {
            for (int j=i*i; j<MAXN; j+=i)
                if (spf[j]==j)
                    spf[j]=i;
        }
    }
}//sieve ends
map<int,int> getfactor(int a)//factor prime
{
    map<int,int> m;
    while(a>1)
    {
        m[spf[a]]++;
        a/=spf[a];
    }
    return m;
}//prime factor ends
int power(int x,int y,int p)
{
    int res = 1;
    x = x%p;
    if (x == 0) return 0;
    while (y > 0)
    {
        if (y & 1)
            res = (res*x)% p;
        y = y>>1;
        x = (x*x)%p;
    }
    return res;
}
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
int inverse(int a, int p)
{
    return power(a, p-2, p);
}
int ncr(int n, int r, int p)
{
    if(r>n)
        return 0;
    if (r==0||r==n)
        return 1;
    return (fac[n]*inverse(fac[r], p) % p *
            inverse(fac[n-r], p) % p) % p;
}
void solve()
{
    int n;
    cin>>n;
    if(n%2)
    {
        int m=1+n/2;
        cout<<2*(m+1)*m<<endl;
    }
    else{
        int m=n/2+1;
        cout<<m*m<<endl;
    }
}
signed main()
{
    fast
    int test=1;
    //cin>>test;
    while(test--)
    {
        solve();
    }
}