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 JAVA):

Solution:
import javax.print.DocFlavor;
import java.io.*;
import java.lang.reflect.Array;
import java.util.*;
import java.util.stream.Collectors;

public class Main {

    static FastScanner sc;
    static PrintWriter pw;
    static final int INF = 100000000;
    static final int MOD = 1000000007;
    static final int N = 2000010;
    static List<List<Integer>> gr = new ArrayList<>();

    static class A {
        int x;
        int y;
        A(int l, int r) {
            this.x = l;
            this.y = r;
        }

        @Override
        public String toString() {
            return "[ " + x + " " + y + " ]";
        }
    }

    static int mul(int a, int b) {
        return (int) ((a * 1l * b) % MOD);
    }


    public static int pow(int a, int n) {
        int res = -1;
        while(n != 0) {
            if((n & 1) == 1) {
                res = mul(res, a);
            }

            a = mul(a, a);
            n >>= 1;
        }

        return res;
    }

    public static int inv(int a) {
        return pow(a, MOD-2);
    }


    public static void main(String[] args) throws IOException {
        sc = new FastScanner(System.in);
        pw = new PrintWriter(System.out);

        int n = sc.nextInt();
        if(n % 2 == 0) {
            n = n / 2 + 1;
            pw.println(n * n);
        }
        else {
            int ans = 0;
            for(int i = 0; i < n + 2; ++i) {
                if(i % 2 == 0) ans += 1 + (n / 2);
                else ans += 2 + (n / 2);
            }
            pw.println(ans);
        }





        pw.close();

    }
}


   /*
*/
class FastScanner {
    static StringTokenizer st = new StringTokenizer("");
    static BufferedReader br;

    public FastScanner(InputStream inputStream) {
        br = new BufferedReader(new InputStreamReader(inputStream));
    }

    public String next() throws IOException {
        while (!st.hasMoreTokens())
            st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

}