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:
#include<bits/stdc++.h>
#define llu unsigned long long  
#define ll long long 
#define l1 int
#define u_m unordered_map
#define M map
#define answer cout<<ans<<endl;
#define Sum cout<<sum<<endl;
#define sort1(s) sort(s.begin(),s.end())
#define pb(x) push_back(x)
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
using namespace std;
/* 
vector<vector<int>> v(101,vector<int>());
vector<bool> b(101,false);
vector<int> ans(101);
 
ll power(ll x, ll y)  
{  
    ll res = 1;        
    while (y > 0)  
    {  
        if (y & 1)  
            res = (res*x)%mod;  
        y = y/2;  
        x = (x*x)%mod;
    }  
    return res%mod;  
}
int dfs(int p,int k)
{
    a[p]=true;
    int i,j=1;
    for(i=0;i<v[p].size();i++)
    {
        if(a[v[p][i]]==false)
        {
            if(k==4)
            {
                ans[v[p][i]]=5;
                a[v[p][i]]=true;
                j=dfs(v[p][i],5);
            }
            else
            {
                ans[v[p][i]]=4;
                a[v[p][i]]=true;
                j=dfs(v[p][i],4);
            }
        }
    }
    return j;
}
ll prime(ll n)
{
    ll mx=n+1;//cout<<mx<<endl;
    for(ll i=2;i*i<=n;i++)
    {
        if(n%i==0 && n/i!=i)
        {
            mx=min(mx,(n/i) + i);
            //cout<<mx<<endl;
        }
    }
    return mx;
}
ll modinv(ll x)
{
    return power(x,mod-2)%mod;    
}
void swap(ll *xp, ll *yp) 
{ 
    ll temp = *xp; 
    *xp = *yp; 
    *yp = temp; 
}
void print(vector<ll> v)
{
    ll i,j,n=v.size();
    for(i=0;i<n;i++)
    {
        cout<<v[i]<<" ";
    }
    cout<<endl;
}
bool sortcol( const vector<int>& v1, 
               const vector<int>& v2 ) { 
 return v1[1] < v2[1]; 
} 
 
bool sortcol1( const vector<int>& v1, 
               const vector<int>& v2 ) { 
 return v1[0] < v2[0]; 
}
void scan(vll &v,ll n)
{
    for(ll i=0;i<n;i++)
    cin>>v[i];
}
ll gcd(ll a, ll b) 
{ 
    if (b == 0) 
        return a; 
    return gcd(b, a % b);  
}
ll lcm(ll a,ll b)
{
    return (a*b)/gcd(a,b);
}
*/ 

int main()
{
    ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
    /*l1 t;
    cin>>t;
    while(t--)*/
    {
    	ll n;
    	cin>>n;
    	if(n==1)
    	{
    		cout<<4;
    		return 0;
    	}
    	if(n%2==0)
    	{
    		n=(n/2 + 1)*(n/2 + 1);
    	}
    	else
    	{
    		n=2*(n/2 + 1)*(n/2 + 2);
    	}
    	cout<<n;
    }
    return 0;
}