//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>

#define ll long long
#define ld long double

#define rep(i, n) for(ll i = 0; i < n; ++i)
#define rep2(i, a, b) for(ll i = a; i <= b; ++i)
#define rrep(i, a, b) for(ll i = a; i >= b; --i)

#define pii pair<int, int>
#define pll pair<ll, ll>

#define fi first
#define se second

#define pb push_back
#define eb emplace_back

#define vi vector<int>
#define vll vector<ll>
#define vpii vector<pii>
#define vpll vector<pll>

#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()

#define endl '\n'
using namespace std;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

const int MOD=1000000007;
//const int MOD=998244353;
const ll INF=9223372036854775807;
const int inf=2147483647;
const double PI=acos(-1);
int dx[8] = {1,0,-1,0,1,1,-1,-1};
int dy[8] = {0,1,0,-1,-1,1,1,-1};
 
const int MAX = 310000;
using Graph = vector<vector<int>>;
Graph G;


signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false); 
  
  ll n;
  cin >> n;
  if(n%2 == 1){
    n++;
    n/=2;
    ll num = n*(n+1)/2;
    num*=4;
    cout << num << endl;
  }
  else{
    n /= 2;
    n++;
    ll num = n*n;
    cout << num << endl;
  }
  return 0;
}
### Prompt

Your challenge is to write a CPP solution to the following 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.

### Response

```cpp
//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>

#define ll long long
#define ld long double

#define rep(i, n) for(ll i = 0; i < n; ++i)
#define rep2(i, a, b) for(ll i = a; i <= b; ++i)
#define rrep(i, a, b) for(ll i = a; i >= b; --i)

#define pii pair<int, int>
#define pll pair<ll, ll>

#define fi first
#define se second

#define pb push_back
#define eb emplace_back

#define vi vector<int>
#define vll vector<ll>
#define vpii vector<pii>
#define vpll vector<pll>

#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()

#define endl '\n'
using namespace std;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

const int MOD=1000000007;
//const int MOD=998244353;
const ll INF=9223372036854775807;
const int inf=2147483647;
const double PI=acos(-1);
int dx[8] = {1,0,-1,0,1,1,-1,-1};
int dy[8] = {0,1,0,-1,-1,1,1,-1};
 
const int MAX = 310000;
using Graph = vector<vector<int>>;
Graph G;


signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false); 
  
  ll n;
  cin >> n;
  if(n%2 == 1){
    n++;
    n/=2;
    ll num = n*(n+1)/2;
    num*=4;
    cout << num << endl;
  }
  else{
    n /= 2;
    n++;
    ll num = n*n;
    cout << num << endl;
  }
  return 0;
}
```