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)

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <fstream>
#include <string>
#include <math.h>
#include <cstdlib>
#include <istream>
#include <sstream>
#include <cctype>
#define fore(i,x,y) for(long long i=x;i<y;i++)
typedef long long ll;
using namespace std;
int main()
{
	ll N,A,B,ANS=0;
	cin >> N >> A >> B;
	ANS=ANS+(N/(A+B))*A;
	if(N%(A+B)<A){
		ANS=ANS+N%(A+B);
	}
	else{
		ANS=ANS+A;
	}
	cout << ANS << endl;
}