How many ways are there to choose two distinct positive integers totaling N, disregarding the order?

Constraints

* 1 \leq N \leq 10^6
* N is an integer.

Input

Input is given from Standard Input in the following format:


N


Output

Print the answer.

Examples

Input

4


Output

1


Input

999999


Output

499999
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author anand.oza
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        ASumOfTwoIntegers solver = new ASumOfTwoIntegers();
        solver.solve(1, in, out);
        out.close();
    }

    static class ASumOfTwoIntegers {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            long n = in.nextLong();

            if (n % 2 == 0)
                out.println(n / 2 - 1);
            else
                out.println(n / 2);
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

    }
}

