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 FOR(i,l,r) for(int i=l;i<=r;i++)
#define ROF(i,l,r) for(int i=l;i>=r;i--)
#define lson k<<1,l,mid
#define rson k<<1|1,mid+1,r
#define r(x) read(x)
#define rr(x,y) read(x);read(y)
#define rrr(x,y,z) read(x);read(y);read(z)
#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
#define fi first
#define se second
#define sss(str) scanf("%s",str+1)
using namespace std;
typedef long long LL;
typedef pair<int,int> pt;
const int N=1e6+5;
const int M=2e3+5;
const int mod=1e9+7;
const int INF=2e9;
const double eps=1e-8;
const double pi=acos(-1);
template <class T>
inline void read(T &x) {
    char c;x = 1;
    while ((c = getchar()) < '0' || c > '9') if (c == '-') x = -1;
    T res = c - '0';
    while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - '0';
    x *= res;
}
int n,m;
char str[N];
int f[N];
void solve(int cas)
{
    r(n);
    int cnt1=n/2,cnt2=n-cnt1;
    LL ans=0;
    set<pt> s;
    for(int i=cnt1;i>=-cnt1;i-=2){
        for(int j=cnt2;j>=-cnt2;j-=2){
            s.insert(mp(i,j));
            s.insert(mp(j,i));
        }
    }
    cout<<s.size()<<endl;
}
int main()
{
    int t;
//    r(t);
    t=1;
//    init();
    FOR(i,1,t){
        solve(i);
    }
    return 0;
}