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: Accepted)

#include<bits/stdc++.h>
  #define f(i,a,b) for(int i=a;i<b;i++)
  #define ll long long
  #define vll vector<long long>
  #define mll map<ll,ll>
  #define pll pair<long long,long long>
  #define pb(x) push_back(x)
  #define F first
  #define S second
  #define print(v) for(auto it:v)cout<<it<<" "
  #define mod 1000000007   //10^9 +7
  #define sorty(v) sort(v.begin(),v.end())
  #define it(v) v.begin(),v.end()
using namespace std;  

int main() {
  ll n,a,b;
  cin>>n>>a>>b;
  ll hold=n/(a+b);
  ll ans=a*hold;
  ll rest=n%(a+b);
  if(rest>=a)ans+=a;
  else ans+=rest;
  cout<<ans;



	return 0;
}