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:
//IaMaNanBord//

#include <bits/stdc++.h>
using namespace std;

//Macro Shorthands
#define F                       first
#define S                       second
#define f(i,n)                  for(ll i=0;i<=n;i++)
#define rf(i,n)                 for(ll i=n;i>=0;i--)
#define Cf(i,a,b)               for(ll i=a;i<=b;i++)
#define Crf(i,b,a)              for(ll i=b;i>=a;i--)
// #define endl                    '\n'           
#define pb                      push_back
#define mp                      make_pair
#define z                       ((ll)1e9 + 7)  
#define every(it,x)             for(auto &it:x)
#define SET(it,x)               for(auto &it:x){cin>>it;}
#define ins                     insert
#define INF						((ll)1e18)
#define Test                    ll T; cin>>T; while(T--)
#define all(v)                  v.begin(),v.end()
#define nline                   cout<<endl
#define SZ(v)                   (ll)v.size()
#define pll                     pair<ll,ll>  
 
//Macro Shorthands
#ifndef ONLINE_JUDGE
    #define revelio(args...)        { string _s = #args; replace(_s.begin(), _s.end(), ',', ' ');stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
#else
    #define revelio(args...)        {}
#endif

void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {cerr << *it << " = " << a << endl; err(++it, args...);}

//Data types
typedef long long               ll;
typedef vector<ll>              vll;
typedef vector<string>          vstr;
typedef vector<char>            vchar;
typedef vector<pair<ll,ll> >    vpll;
typedef set<ll>                 sll;
typedef set<string>             sstr;
typedef set<pair<ll,ll> >       spll;
typedef map<ll,ll>              mllll;
typedef map<string,ll>          mstrll;
typedef queue<ll>               qll;


//Functions
ll powMod(ll x,ll y)            {ll p=1; while(y){if(y%2){p=(p*x)%z;} y/=2; x=(x*x)%z;} return p;}
ll CpowMod(ll x,ll y, ll w)     {ll p=1; while(y){if(y%2){p=(p*x)%w;} y/=2; x=(x*x)%w;}return p;}
ll invMod(ll x)                 {return powMod(x,z-2);}
ll CinvMod(ll x,ll w)           {return CpowMod(x,w-2,w);}
ll gcd(ll a, ll b)              {return b == 0 ? a : gcd(b, a % b);}
// ll exGcd(ll a,ll b,ll &x, ll &y){if(b==0){y = 0;x = 1;return a;}ll xtmp, ytmp;ll g = exGcd(b,a%b,xtmp,ytmp);
//                                 x=ytmp; y = xtmp - (a/b)*ytmp;return g;}
// vll sub;
// void subSet(ll k,ll n)          {if(k==n+1){every(it,sub)cout<<it<<" ";cout<<endl;}
//                                 else{sub.pb(k);subSet(k+1,n);
//                                 sub.pop_back();subSet(k+1,n);}}

// vll perm; const ll N = (ll)2e6 ;bool isSelected[N];
// void permut(ll n)               {if(perm.size()==n){every(it,perm)cout<<it<<" ";cout<<endl;}
//                                 else{Cf(i,1,n){if(isSelected[i])continue;perm.pb(i);
//                                 isSelected[i]=1;permut(n);perm.pop_back();isSelected[i] =0;}}} 




void solve(){
    ll n;
    cin>>n;
    // {sleft,dir,i,j}
    set<pair<pair<ll,char>, pll >> s;
    spll ans;

    // auto dfs = [&](){
    //     s.ins({{1,'r'},{1,0}});
    //     s.ins({{1,'l'},{-1,0}});
    //     s.ins({{1,'u'},{0,1}});
    //     s.ins({{1,'d'},{0,-1}});
    //     while(!s.empty()){
    //         ll m = s.begin()->F.F ; 
    //         char dir = s.begin()->F.S;
    //         ll x = s.begin()->S.F ; 
    //         ll y = s.begin()->S.S ; 
    //         s.erase(s.begin());
    //         if(m==n){
    //             ans.ins({x,y});
    //             continue;
    //         }
    //         if(dir=='r' || dir =='l'){
    //             s.ins({{m+1,'u'},{x,y+1}});
    //             s.ins({{m+1,'d'},{x,y-1}});
    //         }
    //         if(dir=='d' || dir =='u'){
    //             s.ins({{m+1,'r'},{x+1,y}});
    //             s.ins({{m+1,'l'},{x-1,y}});
    //         }
    //     }
    // };
    // dfs();
    // cout<<SZ(ans)/4;
    // nline;
    // 1, 3, 5->6, 7->10, 9->15, 11 -> 21, 13-> 28
    if(n%2){
        cout<<(((n+1)/2+1)*((n+1)/2)/2)*4<<endl;
    }
    else{
        cout<<(n/2+1)*(n/2+1)<<endl;
    }
}


int main(){
    ios_base::sync_with_stdio(0) ;
    cin.tie(0) ; cout.tie(0) ;
    cout.precision(20);

    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
	freopen("error.txt","w",stderr);
    #else
    
    #endif
    solve();
    #ifndef ONLINE_JUDGE
        cout<<"\nTime Elapsed: "<<1.0*clock()/ CLOCKS_PER_SEC<<" Sec\n";
    #endif
    return 0;
}