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;
const int IINF = INT_MAX;
using ll = long long;
using P = pair<int,int>;
#define INF 1000000000000
int n;

int main(){
    ll n,a,b; cin >> n >> a >> b;
    
    ll ans1 = n/(a+b);
    ll ans2 = n%(a+b);
    cout << ans1 << " " << ans2 << endl;
    if(ans2 != 0){
    	if(a == 0){
    		ans2 = 0;
    	} else{
            ans2 = ans2%a;
    	}
    }
    
    cout << ans1*a + ans2 << endl;
    return 0;	
}