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<bits/stdc++.h>
using namespace std;
char s[10000000];
char x[10000000];
char y[10000000];
int main()
{
    long long int n,a,b;
    cin>>s>>x>>y;
    int i;
    i=n=0;
    while(s[i]!='\0')
    {
        n=n*10+s[i]-'0';
        i++;
    }
    i=a=0;
    while(x[i]!='\0')
    {
        a=a*10+x[i]-'0';
        i++;
    }

    i=b=0;
    while(y[i]!='\0')
    {
        b=b*10+y[i]-'0';
        i++;
    }
    int c=0;
    while(n>=0)
    {
        if(c>0&&a>n)
        {
            a+=n;
        }
        n=n-a;
        n=n-b;
        c++;
    }

    /*while(a>=(n-(b+a)))
    {
        cout<<a+(n-(b+a));
    }*/
    cout<<a;
    cout<<endl;


    //cout<<n<<" "<<a<<" "<<b;

    return 0;
}