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>
#define rep(i,n) for(int i = 0;i < n;i++)
using namespace std;
typedef long long ll;

int main(){
    ll n,a,b;
    cin >> n >> a >> b;
    ll nn =n;
    ll cnt  =0;
    ll j = (a+b)/n;

    if((a+b) >= n) cout << a;
    else{
        cnt = a*j;
        n -= cnt;
        while(true){
            n -= a;
            cnt += a;
            if(n <= 0){
                cout << n + cnt;
                break;
            }
            n-=b;
            if(n <= 0){
                cout << cnt;
                break;
            }
        }
    }

    return 0;
}