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: Wrong Answer)

#include<stdio.h>
#define ll long long int
int main(){
	ll n,a,b;
	scanf("%lld%lld%lld",&n,&a,&b);
	if(a==0)
		printf("0");
	if(a+b==n&&a!=0)
		printf("%lld",a);
	if(b==0&&a!=0)
		printf("%lld",n);
	if(a+b<n&&a!=0){
		ll c;
		c=n/(a+b);
		ll d;
		d=n%(a+b);
		if(d>a){
			c++;
			c=c*a;
			}
		else{
			c=c*a;
			c+=d;
		}
		printf("%lld",c);	
	}
	
	
}