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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <utility>
#include <tuple>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <cmath>
#include <iomanip>
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) (int)(x).size()
using namespace std;
using ll = long long;
constexpr ll INF = 3000000000000000000;

int main() {
    int X, Y, Z;
    cin >> X >> Y >> Z;
    cout << (X - Z) / (Y + Z) << "\n";
    return 0;
}