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: Accepted)

/*
	Copyright: razouq (c)
	Author: Anass BENDARSI
	Date: 07-Mar-20 1:06:41 PM
	flamers will **** you
*/
//#include<flamers.h>

#include "bits/stdc++.h"
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#define ll long long
#define ull unsigned long long
#define F first
#define S second
#define PB push_back
#define POB pop_back

using namespace std;

int main(){
//	freopen("input.in", "r", stdin);
//	freopen("output.out", "w", stdout);
	ios::sync_with_stdio(0);
	cin.tie();
	long long n, a, b;
	cin>>n>>a>>b;
	ll res = a * (n / (a + b));
	res += min((n % (a + b)), a);
	cout<<res<<endl;
	return 0;
}