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 <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdio.h>
int compare(const void *a,const void *b){
    if(*(int *)a>*(int *)b)return 1;
    return -1;
}
long long MAX(long long a,long long b){
    return a>b?a:b;
}
int main()
{   long long n,a,b;
    scanf("%lld%lld%lld",&n,&a,&b);
    long long cnt=0,ans=0,add=0;
    while(1){
        if(add==0&&cnt+a<=n){
            ans+=a;
            cnt+=a;
            add=1;
        }else if(add==0){
            printf("%lld",MAX(ans+(n-cnt),0));
            return 0;
        }else{
            add=0;
            cnt+=b;
        }
    }
    
    return 0;
}