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)

#include<stdio.h>
#include <stdlib.h>
#include<string.h>
#include<math.h>
#define loop(i, l, r) for(i=l; i<r; i++)/*Instead of function, (for).*/
#define rep(i,r) for(i=0; i<r; i++)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long int lli;
/*Return Combination ,(nCr)*/
int combi(int n, int r)
{
	if(r==0 || r==n) return 1;
	else return combi(n-1,r) + combi(n-1,r-1);
}
/*Return gcd(a, b)*/
int gcd(int a, int b){
	if(b == 0) return a;
	else return gcd(b, a % b);
}
int main(void)
{
	lli n, a, b, c, d, ans;
	scanf("%lld%lld%lld", &n, &a, &b);
	if(a==0) ans = 0;
	c= n/(a+b);
	d = n%(a+b);
	if(0<=d && d<a) ans = c*a+d;
	if(a<=d && d<a+b) ans = c*a+a;
	printf("%lld", ans);
	return 0;
}