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>
#include<stdlib.h>
#include<math.h>
int main(){
  int unused __attribute__((unused));
  int N, A, B, ans;

  unused = scanf("%d %d %d", &N, &A, &B); //N個のボール，A:青ボール，B:赤ボール

  if(A==0) printf("%d", 0);
  else if(N<=A) printf("%d", N);
  else if(N<=(A+B)) printf("%d", A);
  else if(A>0 && B>0){
    ans = (N/(A+B))*A;

    if(N%(A+B) > A) ans += A;
    else ans += N%(A+B);
    printf("%d", ans);
  }

  return 0;
}