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: Time Limit Exceeded)

#include <iostream>
#include <stdio.h>
#include <vector>
#include <algorithm>
#include <math.h>
#include <numeric>
#include <iomanip>
#include <map>

#define REP(i, n ,N) for(ll i = (n); i < (N); i++)
//#define RREP(i, n ,N) for(ll i = (N-1); i >= (n); i--)
#define p(s) cout<<(s)<<endl
#define p2(a, b) cout<<(a)<<" "<<(b)<<endl

using namespace std;
typedef long long  ll;

const long long MOD = 1e9+7;

int main(){

    ll n, a, b;
    cin >> n >> a >> b;

    ll count = 0;
    ll ans = 0;

    while(count <= n){
        ans += a;
        count += a + b;
    }

    if(count - n > b){
        ans -= count - n - b;}

    p(ans);

    return 0;
}