For a positive integer n let's define a function f:

f(n) = - 1 + 2 - 3 + .. + ( - 1)nn

Your task is to calculate f(n) for a given integer n.

Input

The single line contains the positive integer n (1 ≤ n ≤ 1015).

Output

Print f(n) in a single line.

Examples

Input

4


Output

2


Input

5


Output

-3

Note

f(4) = - 1 + 2 - 3 + 4 = 2

f(5) = - 1 + 2 - 3 + 4 - 5 = - 3
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Scanner;

/**
 * 题意：f(n) =  -1+2-3+4-5+6-7... 。f(4) =  - 1 + 2 - 3 + 4 = 2 ; f(5) =  - 1 + 2 - 3 + 4 - 5 =  - 3
 *
 * @author TinyDolphin
 *         2017/7/3 15:55.
 */
public class Main {

    public static void main(String[] args) throws IOException {
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        BigInteger inputN;
        BigInteger ans;
        while (in.hasNext()) {
            inputN = in.nextBigInteger();
            //判断一个数是否是偶数：inputN & 1 == 0 时，为偶数。如：4==> 0100 & 0001 = 0000 ；5==> 0101 & 0001 = 0001
            //ans = inputN & 1 == 0 ? inputN / 2 : -(inputN / 2 + 1);
            //下面为转化为大整数进行运算
            ans = inputN.and(BigInteger.valueOf(1)).compareTo(BigInteger.valueOf(0)) == 0 ? inputN.divide(BigInteger.valueOf(2)) : inputN.divide(BigInteger.valueOf(2)).add(BigInteger.valueOf(1)).negate();
            out.println(ans);
        }
        out.flush();
    }
}