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: Accepted)

// AtCoder150 C

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <math.h>
#include <assert.h>
#include <string.h>
#include <memory.h>
#include <malloc.h>
#include <stdlib.h>
#include <ctype.h>

// qsortで用いる比較用関数．*a<*bならaが先になる．
int compare(const void* a, const void* b) {
    return *(int*)a - *(int*)b;
}

int revcompare(const void* a, const void* b) {
    return *(int*)b- *(int*)a;
}

int main(void)
{
    int i,j,n;
    long long int N,a,b;
    long long int cnt = 0;

    scanf("%lld %lld %lld", &N, &a, &b);

    cnt += (N / (a+b)) * a;
    if(N %(a+b) > a){
        cnt += a;
    }else{
        cnt += N %(a+b);
    }
    printf("%lld", cnt);
    return 0;

}