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>
using namespace std;

int main() {
    long long n,a,b;
    cin>>n>>a>>b;
    if (a==0) cout<<0<<endl;
    else if (a>=n) cout<<n<<endl;
    else if (a+b>=n) cout<<a<<endl;
    else {
        long long ans = floor(n/(a+b)),ans1=0,ans2=0;
        ans2=ans*a;
        ans1=ans*a+ans*b;
        if(ans1+a<=n){
            ans1+=a;
            ans2+=a;
        }
        else ans2+=(n-ans1);
        cout<<ans2<<endl;
        
    }
        
}