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)

#define _CRT_SECURE_NO_WARNINGS
#include"stdio.h"
#include<string.h>

int main() {
	long int red=0;
	long int blue=0;
	long int btotal=0;
	long int ired;
	long int iblue;
	long int itotal;
	scanf("%ld %ld %ld", &itotal, &iblue, &ired);
	long int i;
	long int cnt = 0;
	//printf("%d", itotal);

	long int calc;
	int nokori;
	calc = iblue + ired;
	nokori = itotal % calc;
	long int full;
	full = itotal - nokori;
	long int times;
	times = full / calc;


	for (i = 0; i < calc; i++) {
		if (cnt == 0) {
			if (blue == iblue) {
				cnt = 1;
				blue = 0;
				//break;
			}
			else {
				blue++;
				btotal++;
			}

			if (blue == iblue) {
				cnt = 1;
				blue = 0;
				//break;
			}
		}
		else {
			if (red == ired) {
				cnt = 0;
				red = 0;
				//break;
			}
			red++;
			if (red == ired) {
				cnt = 0;
				red = 0;
				//break;
			}

		}
		//printf("a", btotal);


	}
	long int p1;
	p1 = btotal * times;
	btotal = 0;
	blue = red = 0;

	for (i = 0; i < nokori; i++) {
		if (cnt == 0) {
			if (blue == iblue) {
				cnt = 1;
				blue = 0;
				//break;
			}
			else {
				blue++;
				btotal++;
			}

			if (blue == iblue) {
				cnt = 1;
				blue = 0;
				//break;
			}
		}
		else {
			if (red == ired) {
				cnt = 0;
				red = 0;
				//break;
			}
			red++;
			if (red == ired) {
				cnt = 0;
				red = 0;
				//break;
			}

		}
		//printf("a", btotal);


	}

	printf("%ld", p1 + btotal);
}