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>
#include <stdlib.h>
#include <string.h>

int main(){
    unsigned long long n,a,b,count,mod;
    char token[100]={};
    while(fgets(token, sizeof(token), stdin)!=NULL){
        n=atoi(strtok(token, " "));a=atoi(strtok(NULL, " "));b=atoi(strtok(NULL, " "));
        if ((a+b)==0) {
            printf("0");
            return 0;
        }
        count=(n/(a+b))*a;
        mod=n%(a+b);
        if(mod<=a) count+=mod;
        else count+=a;
        printf("%lld",count);
        return 0;
    }
    return 0;
}