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>
using namespace std;


#define mod 1000000007
#define float double
#define min3(a, b, c) min(c, min(a, b))
#define min4(a, b, c, d) min(d, min(c, min(a, b)))
#define rrep(i, n) for(int i=n-1;i>=0;i--)
#define fast ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
#define gc getchar_unlocked
#define fo(i, n) for (int i = 0; i < n; i++)
#define rep(i,a,n) for (int i = a; i < n; i++)
#define Fo(i, k, n) for (i = k; k < n ? i < n : i > n; k < n ? i += 1 : i -= 1)
#define li long int
#define int long long int
#define uli unsigned long int
#define ull unsigned long long int
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define mem(x) memset(x, -1, sizeof(x))
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for (auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
#define INF 9223372036854775807
#define endl "\n" 
typedef pair<int, int>	pii;
//typedef pair<ll, ll>	pl;
typedef vector<int>		vi;
//typedef vector<ll>		vl;
typedef vector<pii>		vpii;
//typedef vector<pl>		vpl;
typedef vector<vi>		vvi;
//typedef vector<vl>		vvl;
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());


bool isPrime(int n){
    if(n==1) return false;
    if(n==2) return true;
    for(int i=2;i*i<=n;i++){
        if(n%i==0)return false;
    }
    return true;
}

bool sortbysec(const pair<int,int> &a,const pair<int,int> &b){
        return (a.second > b.second);
}

bool sortbycompare(int a,int b){
        return (abs(a) > abs(b));
}


int32_t main(){
fast
int t=1;
// cin>>t;
while(t--){
int n, m, x, y, z, i, j, k = 0, l = 0, mn=LLONG_MAX,mx=LLONG_MIN;
    cin>>n;
    if(n%2==0){
        n=n/2;
        cout<<(n+1)*(n+1);
    }
    else{
        n=n/2;
        cout<<2*(n+1)*(n+2);
    }

}
return 0;
}