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>
int main(void)
{
    int je=0;
    long a,b,n,s,cou;
    scanf("%ld%ld%ld",&n,&a,&b);
    
    s = (long)a + b;
    cou = 0;
    if((a + b) <= n){
        while((n - b) >= 0){
            if(je == 0){
                n = n - a;
                cou++;
                je = 1;
            }
            else if(je == 1){
                n = n - b;
                je = 0;
            }
        }
        cou = cou * a + n;
    }
    else if(a >= n)
        cou = n;
    else{
        cou = a;
    }
    printf("%ld\n",cou);
    return 0;
}