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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
 
 
void swap (int *x, int *y) {
  int temp;    
 
  temp = *x;
  *x = *y;
  *y = temp;
}
 
int gcd(int a,int b){
  if(a<b){
    swap(&a,&b);
  }
 
  if(b==0){
    return a;
  }
 
  int r=a%b;
  while(r!=0){
    a = b;
    b = r;
    r = (a%b);
  }
 
  return b;
}
 
/* クイックソート */
int dn(const void*a,const void*b){return*(int*)b-*(int*)a;}
 
 
int modpow(int a,int n,int mod){
  int ans=1;
  while(n>0){
    if(n & 1){
      ans=ans*a%mod;
    }
    a=a*a%mod;
    n/=2;
  }
  return ans;
}
 
int max(int a,int b){
  if(a<b){
    return b;
  }
  else{
    return a;
  }
}
 
int min(int a,int b){
  if(a<b){
    return a;
  }
  else{
    return b;
  }
}
 
void chmax(int *a,int b){
  if(*a < b){
    *a = b;
  }
 
  return;
}
 
void chmin(int *a,int b){
  if(*a > b){
    *a =b;
  }
 
  return;
}
 

int main(void){
  long int n,a,b;
  scanf("%ld%ld%ld",&n,&a,&b);
  long int count=(n+a+b-1)/(a+b);
  long int mod=n%(a+b);
  long int ans=(count-1)*a;
  if(mod==0 || b<=mod){
    ans+=a;
  }else{
    ans+=mod;
  }
  printf("%ld\n",ans);
  return 0;
}