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; 

#define ll long long
#define ar array
// long a,b,n,d;
// long ans = 0;
// void solve(){
//     cin >> n;
//     long x = n/2 ;
//     while(x >= 0){
//         ans += (n-1)*4*x;
//         n-=2 ;
//         --x;
//     }
//     cout << ans ;


// }
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    ll n,a,b;
    cin >> n >> a>> b;
    ll c = n%(a+b);
    if(c >=a ){
        cout << n/(a+b)*a+a;
    }else{
        cout << n/(a+b)*a + c;
    }

}