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:

/*
Name:-Kautik Khokhar
ID:-201901064
*/
#include <bits/stdc++.h>
using namespace std;

#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define inarr(a,n) for(int i=0;i<n;i++) cin>>a[i]
#define outarr(a,n) for(int i=0;i<n;i++) cout<<a[i]<<" "; cout<<endl
#define maxa(a,n) *max_element(a,a+n)
#define mina(a,n) *min_element(a,a+n)
#define Max(x,y,z) max(x,max(y,z))
#define Min(x,y,z) min(x,min(y,z))
#define fr(i,s,e) for(int i=s;i<e;i++)
#define rf(i,s,e) for(int i=s-1;i>=e;i--)
#define rsort(a,n) sort(a,a+n,greater<int>())
#define cp(m) cout<<setprecision(m)
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define outvect(s) for(auto x:s) cout<<x<<" ";

typedef long long ll;
typedef long double ld;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef vector<pair<int,int> > vpi;

#define PI 3.141592653589793
#define MOD 1000000007
bool sortbysec(const pair<int,int> &a,const pair<int,int> &b)
{
    return (a.second < b.second);
}
ll gcd(ll a,ll b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);

}
ll pox(ll n,ll x)
{
    int ans=0;
    while(n%x==0)
    {
        ans++;
        n/=x;
    }
    return ans;
}
ll s2n(string s)
{
    stringstream geek(s);
    ll x = 0;
    geek >> x;
    return x;
}
string n2s(ll n)
{
    string s=to_string(n);
}
string revers_string(string s)
{
    string s1=string(s1.rbegin(),s1.rend());
    return s1;
}
int N=10000;
bool prime[10000];
void sieveOfEratosthenes()
{
    fr(i,0,10000)
    prime[i]=true;
    for (int i = 2; i <= sqrt(10000); ++i) {
        if (prime[i]) {
            for (int j = i * i; j <=10000; j += i) {
                prime[j] = false;
            }
        }
    }
}
int main()
{
	IOS;
    int n;
    cin>>n;
    cout<<(n/2+1)*((1-n%2)*(n/2+1)+(n%2)*(2*(n/2+2)));
	return 0;
}