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: Time Limit Exceeded)

#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <math.h>

//#define MOD (int)(1e5 + 7)
#define MOD (100007)

static inline void done(const char *s)
{
	puts(s);
	exit(0);
}

#define yes() done("Yes");
#define no() done("No");
#define invalid() done("-1");

static int comp(const void *a, const void *b)
{
	int *pa = (int *)a;
	int *pb = (int *)b;
	return *pa < *pb;
}

static inline int *readNint(int n)
{
	int *a = calloc(sizeof(int), n);
	for (int i = 0; i < n; i++)
		scanf("%d", &a[i]);
	return a;
}

static inline long *readNlong(int n)
{
	long *a = calloc(sizeof(long), n);
	for (int i = 0; i < n; i++)
		scanf("%ld", &a[i]);
	return a;
}

int main(int argc, char **argv)
{
	uint64_t N, A, B;
	scanf("%ld%ld%ld", &N, &A, &B);
	uint64_t X = 0;
	uint64_t Y = 0;
	while (X < N) {
		Y += A;
		X += A;
		if (X >= N) {
			uint64_t r = X - N;
			Y -= r;
			break;
		}
		X += B;
	}
	printf("%ld\n", Y);
	return 0;
}