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(){
unsigned long N,A,B;
unsigned long sum = 0;
  scanf("%lu %lu %lu",&N,&A,&B);
//  printf("%d",N);
unsigned long  int D = N / (A + B);
unsigned long  int D1 = N % (A + B);

  if (D1 == 0){
   	printf("%lu",A*D);
  }
  else{
	  for(int i = 0; i < D1;i++){
  		if(i%(A+B) < A ) sum++; 
 	 }  
 	 printf("%lu",sum + (D)*A);
  }
    return 0;
}