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: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <string>
#include <math.h>
#include <stdint.h>
#include <queue>
#include <bitset>
#include <iomanip>
#include <vector>
#include <map>
 
#define PI (3.14159265358979323846)
#define INF (1e+9)
#define DV_NUM (100000007)
using namespace std;

int main() {
    int N,A,B;
    map<int,int,int>q;
    cin>>N>>A>>B;
    int S = N/(A+B);
    int t = N - (A+B)*S;
    if (t <= A){
        cout << S*A+t<< endl;
    }
    else{
        cout << (S+1)*A<< endl;
    }
    
    return 0;
}