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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
int main()
{
    long int a,b,c,d;
    long int i,j,k=0,l,n;
    scanf("%ld%ld%ld",&n,&b,&a);
    if(b==0 || b+a==n || (b+a>n && n>b))
    {
        printf("%ld\n",b);
    }
    else if(b+a>n && b>n)
    {
        printf("%ld\n",n);
    }
    else if(b+a<n)
    {
        k=0;
        for(;n>0;)
        {
            k=k+b;
            n=n-(a+b);
            if(n>b)
                b=b;
            else if(n<b)
                b=n;
        }
        printf("%ld\n",k);
    }
    return 0;
}