Problem:
Score : 200 points

Problem Statement

Takahashi has many red balls and blue balls. Now, he will place them in a row.

Initially, there is no ball placed.

Takahashi, who is very patient, will do the following operation 10^{100} times:

Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.

How many blue balls will be there among the first N balls in the row of balls made this way?

Constraints

1 \leq N \leq 10^{18}

A, B \geq 0

0 < A + B \leq 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

Print the number of blue balls that will be there among the first N balls in the row of balls.

Sample Input 1

8 3 4

Sample Output 1

4

Let b denote a blue ball, and r denote a red ball. The first eight balls in the row will be bbbrrrrb, among which there are four blue balls.

Sample Input 2

8 0 4

Sample Output 2

0

He placed only red balls from the beginning.

Sample Input 3

6 2 4

Sample Output 3

2

Among bbrrrr, there are two blue balls.

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define all(v) v.begin(), v.end()
using in = int64_t;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
const double PI=3.14159265358979323846;
const in MOD = 1000000007;
const in INF=1<<30;
using P=pair<int,int>;
vector<int> dx={0,1,-1,0};
vector<int> dy={1,0,0,-1};



int main(){
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
cout << fixed << setprecision(10);
    int n,a,b;
    cin>>n>>a>>b;
    cout<<a*(n/(a+b))+n-(a+b)*(n/(a+b))<<endl;
}