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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>

int main(void){
    int n,a,b,sum=0,bc=0,rc=0,ans,f,d;

    scanf("%d %d %d",&n,&a,&b);

    while(true){
        sum=sum+a;
        bc++;
        if(sum>=n){
            f=1;
            d=n-(sum-a);
            break;
        }
        sum=sum+b;
        rc++;
        if(sum>=n){
            f=0;
            break;
        }
    }

    if((bc*a+rc*b)>n){
        if(f==1){
            printf("%d\n",(a*(bc-1)+d));
        }
        if(f==0){
            printf("%d\n",a*bc);
        }
    }

    if((bc*a+rc*b)==n){
        printf("%d\n",a*bc);
    }

    return 0;
}