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;
using ll = long long;
using P = pair<int,int>;

int main() {
	ll n; cin >> n;
    ll a, b; cin >> a >> b;
    int count = 0, total = 0;
    total = a + b;
    if (total == 0) {
      cout << 0 << endl;
      return 0;
    }
    ll alt = n, ans = 0;
    count = alt / total;
    //cout << count << endl;
    //cout << tmp << endl;
    ans = count * a;
    //cout << ans << endl;
    ll rest = n % total;
    if ( rest < a ) {
      ans += rest;
    }
    else ans += a;
    cout << ans << endl;
    
    return 0;
    
}