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: Accepted)

#include<stdio.h>
int main(){
    long n,a,b;
    long sum;
    long x,amari;
    long flag = 0;
    
    scanf("%ld %ld %ld",&n,&a,&b);
    sum = a+b;
    
    x = n/sum;
    amari = n % sum;
    
    if(sum >= n && n <= a) printf("%ld\n",n);
    else if(sum >= n && n > a) printf("%ld\n",a);
    else flag = 1; //ans = a*x;
    
    if(amari > a && flag == 1) printf("%ld\n",(a*x)+a);
    else if(amari <= a && flag == 1) printf("%ld\n",(a*x)+amari);
    else
        
    return 0;
}