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 long N,A,B;
    int flg=0;//0_blue 1_red
    scanf("%lld %lld %lld",&N,&A,&B);
    long long count=0;
    while(1){
        if(flg==0){
            N=N-A;
            flg=1;
            count+=A; 
        }else{
            N=N-B;
            flg=0;
            if(N<=0){
                printf("%lld\n",count);
                return 0;
            }
        }
        if(N<=0){
            count+=N;
            break; 
        }
    }
    printf("%lld\n",count);
    return 0;
}