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: Runtime Error)

#include <stdio.h>

int main()
{
    int r, b, N, i, jr, jb, k, d, ini=0;
    int sum=0, ib;
    static int str[100*100*100*100];
    
    scanf("%d %d %d", &N, &ib, &r);
    
    if (r+ib==0) {
        printf("%d", 0);
    }else{
    
    d=1+N/(ib+r);
    
    b=ib;
    
    for(i=0; i<d; i++){
        for(jb=ini; jb<b; jb++){
            str[jb]=1;
        }
        for(jr=b; jr<b+r; jr++){
            str[jr]=0;
        }
        ini=r+b;
        b=b+(r+ib);
    }
    
    for(k=0; k<N; k++){
        sum=sum+str[k];
    }
    
    printf("%d", sum);
    }
    
    return 0;
}