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.
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pb push_back
#define all(x) x.begin(), x.end()
#define sortall(x) sort(all(x))
#define reverseall(x) reverse(all(x))
#define to_upper(x) transform(x.begin(), x.end(), x.begin(), ::toupper)
#define to_lower(x) transform(x.begin(), x.end(), x.begin(), ::tolower)
#define FOR(i,n) for(int i=0;i<n;i++)
#define FORS(i,s,n) for(int i=s;i<n;i++)
#define send {ios_base::sync_with_stdio(false);}
#define help {cin.tie(NULL); cout.tie(NULL);}
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "= " << x << ", " << #y << "= " << y << endl
#define clr(x) memset(x, 0, sizeof(x))
#define PI 3.1415926535897932384626
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;
typedef priority_queue<int, vi, greater<int>> minpq;
typedef priority_queue<ll, vi, greater<ll>> minpql;
const int mod = 1000000007;
void YES(bool t = 1) { const string YESNO[2] = {"NO", "YES"};cout << YESNO[t] << endl; }
void Yes(bool t = 1) { const string YesNo[2] = {"No", "Yes"};cout << YesNo[t] << endl; }
void yes(bool t = 1) { const string yesno[2] = {"no", "yes"};cout << yesno[t] << endl; }
inline bool isPowerOfTwo(int x) { return x && (!(x&(x-1))); } 
ll mpow(int b, int e, int modu=0) {
    ll result = 1,base = b, exp = e;
    if(modu) base %= mod;
    while (exp > 0) {
        if (exp & 1){
            result = ((ll)result * base) ;
            if(modu) result %= mod;
        }
        base = ((ll)base * base);
        if(modu) base %= mod;
        exp >>= 1;
    }
    return result;
}
int64_t bpow(int b, int e){
    int64_t result = 1,base = b, exp = e;
    while (exp > 0) {
        if (exp & 1){
            result = ((int64_t)result * base) ;
        }
        base = ((int64_t)base * base);
        exp >>= 1;
    }
    return result;
}
bool isPerfectSquare(long double x){long double sr = sqrt(x); return ((sr - floor(sr)) == 0);} 
bool isIn(string &s2, string &s1){if (s1.find(s2) != string::npos) return true;return false;}
bool isSorted(vi &arr){for(int i=0;i<arr.size()-1;i++) if(arr[i] > arr[i+1]) return false;return true;}
void print_v(vi &arr){for(auto x: arr) cout << x <<',';cout << "\b \n";}

int TEST_CASES = 0;
void submain(){
    int n;
    cin >> n;
    int k = n/2;
    ll ans = 0;
    if(n&1){
        ans = (k+1)*(k+2) + (k+2)*(k+1);
    }else{
        ans = (k+1)*(k+1);
    }
    cout << ans << '\n';
    

}

int main()
{
	// ios_base::sync_with_stdio(false);
	// cin.tie(NULL);
    // freopen("in.txt", "w", stdout);
    send help
    if(TEST_CASES){
        int t;
        cin >> t;
        while(t--){
            submain();
        }
    }else{
        submain();
    }
    
	return 0;
    
}
