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)

//#pragma GCC optimize("-O3")
/*
				C++
				encoding: UTF-8
				Last Modified: 07/Mar/20 05:35:24 PM
*/

#include <bits/stdc++.h>
#ifdef H4wk3ye
#	include <prettypr.hpp>
#endif

using namespace std;

// clang-format off

#define		int                    long long
#define		ve                     vector
#define		pb                     push_back
#define		endl                   "\n"
#define		ff                     first
#define		ss                     second
#define		pii                    pair<int, int>
#define		len(v)                 (int)v.size()
#define		all(v)                 v.begin(), v.end()
#define		reset(a, b)            memset(a, b, sizeof(a));
#define		fr(i, s, n)            for (int i = s ; i < n ; ++i)
#define		dfr(i, s, n)           for (int i = s ; i > n ; --i)

template < typename T > void pr (const T& t) {
    cout << t << "\n";
}

template < typename T, typename U, typename... ARGS >
void pr (const T& t, const U& u, const ARGS&... args) {
    cout << t << " ";
    pr (u, args...);
}

// clang-format on


signed main () {
	ios_base::sync_with_stdio (false), cin.tie (nullptr);
	int n, a, b;
	cin >> n >> a >> b;
	int t = a + b;
	int ans = a * (n / t);
	int p = n - (n / t) * t;
	ans += min (a, p);
	pr (ans);
	return 0;
}