Problem:
Score : 200 points

Problem Statement

We have a long seat of width X centimeters.
There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters.

We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person.

At most how many people can sit on the seat?

Constraints

All input values are integers.

1 \leq X, Y, Z \leq 10^5

Y+2Z \leq X

Input

Input is given from Standard Input in the following format:

X Y Z

Output

Print the answer.

Sample Input 1

13 3 1

Sample Output 1

3

There is just enough room for three, as shown below:

Figure

Sample Input 2

12 3 1

Sample Output 2

2

Sample Input 3

100000 1 1

Sample Output 3

49999

Sample Input 4

64146 123 456

Sample Output 4

110

Sample Input 5

64145 123 456

Sample Output 5

109

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
    static int[][] map;
    static int[][] label;


    static ArrayList<String> list;

    static int M;
    static int N;
    static int T;
    static int P;

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int X = scanner.nextInt();
        int Y = scanner.nextInt();
        int Z = scanner.nextInt();


        int temp = X - Z;

        System.out.println(temp / (Y + Z));

    }



}