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:
/*

    SINISTER RULES

*/
#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '"' << x << '"';}
void __print(const string &x) {cerr << '"' << x << '"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

#define sort(arr) sort(arr.begin(),arr.end());
#define reverse(arr) reverse(arr.begin(),arr.end());

vector<int> input(int n)
{
    vector<int> a(n);
    for(int i=0;i<n;++i)    cin>>a[i];
    return a;
}
vector<lli> input_lli(int n)
{
    vector<lli>a(n);
    for(int i=0;i<n;++i)    cin>>a[i];
    return a;
}
void print(vector<int> arr)
{
    for(int i=0; i<arr.size(); ++i)
        cout<<arr[i]<<' ';
    cout<<endl;
}
lli gcd(lli a,lli b)
{
    if(b==0)    return a;
    return gcd(b, a%b);
}
int directions(int n, int uporside)
{
    map<pair<int, int>,int>dp;
    dp[make_pair(0,0)]=1;
    for(int i=0;i<n;++i)
    {
        map<pair<int,int>,int>l;
        //north
        if(i%2==uporside)  
        {for(auto x:dp)
        {
            int y = x.first.second+1;
            l[make_pair(x.first.first,y)]=1;
        }
        //south
        for(auto x:dp)
        {
            int y = x.first.second-1;
            l[make_pair(x.first.first,y)]=1;
        }}
        else
        {//east
        for(auto x:dp)
        {
            int xx = x.first.first-1;
            l[make_pair(xx,x.first.second)]=1;
        }
        //west
        for(auto x:dp)
        {
            int xx = x.first.first+1;
            l[make_pair(xx,x.first.second)]=1;
        }}
        dp = l;
        // debug(dp);
    }   
    debug(dp.size());
    return dp.size();
}


int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;cin>>n;
    // int a = directions(n, 0);
    // if(n%2==0)  cout<<a<<"\n";
    // else    cout<<2*a<<'\n';
    if(n%2==0)
    {
        cout<<lli(pow(n-(n/2-1),2))<<endl;
    }
    else
    {
        cout<<(n+1)*(n-(n/2-1))<<endl;
    }

    return 0;
}