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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int main()
{
   int n, a ,b, c=0, cntball=0;
   string s="";
   cin>>n>>a>>b;
   for(int i=0 ; i<n ; i++){
       if(cntball==n)break;
    for(int j=0 ; j<a ; j++){s+='b';cntball++; if(cntball==n)break;}
    if(cntball==n)break;
    for(int k=0 ; k<b ; k++){s+='r';cntball++; if(cntball==n)break;}
    if(cntball==n)break;
   }
   for(int i=0 ; i<s.size() ; i++)
   if(s[i]=='b')c++;
   
   cout<<c<<" "<<s;;
}