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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<string>
#include<cstring>
#include<iomanip>
using namespace std;
typedef long long lng;
typedef unsigned long long ulng;

#define pb push_back
#define SZ(a) int((a).size())
#define ALL(a) (a).begin(), (a).end()
#define FOR(i,a,b) for(int i=(a); i<=(b); ++i)
#define FORD(i,b,a) for(int i=(b); i>=(a); --i)
#define REP(i,n) FOR(i,0,n-1)
#define UNIQUE(a) a.erase(unique(ALL(a)), a.end()) // <a> must be sorted in advance

lng power(lng b, lng e) {lng sol=1; while(e>0) {if(e&1) {sol=sol*b;} e>>=1; b*= b;} return sol;}


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int X, Y, Z;
    cin >> X >> Y >> Z;

    int x = X - Z;
    cout << x / (Y+Z) << endl;
}