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>
#define ll long long

int main(void) {
    ll N, A, B, ab, remainder, b;
    scanf("%lld %lld %lld", &N, &A, &B);
    if (A == 0) {
        printf("0\n");
        return 0;
    } else if (B == 0) {
        printf("%lld\n", N);
        return 0;
    }
    ab = A + B;
    remainder = N % ab;
    if (remainder > 0) {
        ll tmp = N - remainder;
        b = (tmp * A) / ab;
        remainder -= A;
        b += A;
        if (remainder < 0) {
            b += remainder;
        }
    } else {
        b = (N * A) / ab;
    }
    
    printf("%lld\n", b);
    return 0;
}