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 f first
#define s second
#define fore(i,a,b) for(int i = (a), ThxMK = (b); i < ThxMK; ++i)
#define pb push_back
#define all(s) begin(s), end(s)
#define _ ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sz(s) int(s.size())
#define ENDL '\n'
#define vv(type, name, h, ...) vector<vector<type>> name(h, vector<type>(__VA_ARGS__))
#define vvv(type, name, h, w, ...) vector<vector<vector<type>>> name(h, vector<vector<type>>(w, vector<type>(__VA_ARGS__)))
using namespace std;
template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;
typedef long long lli;
typedef pair<int,int> ii;
typedef vector<int> vi;
#define deb(x) cout << #x": " << (x) << endl;

lli gcd(lli a, lli b){return (b?gcd(b,a%b):a);}
lli lcm(lli a, lli b){ if(!a || !b) return 0; return a * b / gcd(a, b); }
int popcount(lli x) { return __builtin_popcountll(x); }

// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// int rnd(int n){return uniform_int_distribution<int>(0, n-1)(rng);}

lli poww(lli a, lli b){
	lli res =1;
	while(b){ if(b&1) res = res * a; a = a*a; b/=2; }
	return res;
}

vvc<int> graph(int n, int m, bool dir=1){
	vv(int,v,n+1,0);
	fore(i,0,m){
		int a,b; cin>>a>>b;
		v[a].pb(b);
		if(dir)v[b].pb(a);
	}
	return v;
}
template <typename T> static constexpr T inf = numeric_limits<T>::max() / 2;
// ---- コーディングはここから！ ('-')7


void solve(){
	int n; cin>>n;
	vc<lli> dp(2*n+1,0);
	fore(i,0,n)dp[i*2+1]=(i+1)*(i+2)*2;
	fore(i,1,n+1)dp[i*2]=(i+1)*(i+1);
	cout<<dp[n]<<ENDL;

}

int main(){_
	//int t; cin>>t; while(t--)
	solve();
}