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

#define rep(i,n) for (ll i = 0; i < (n); i++)
#define ru(a,b) (((a)+((b)-1)) / (b))   //切り上げ(a/b)
#define dump(x)  cout << #x << " = " << (x) << endl;

int main()
{
	ll n, a, b;
	cin >> n >> a >> b;
	ll ans = 0;	//青の数

	if (a == 0) {
		cout << ans << endl;
		return 0;
	}

	//while (1) {	//操作

	//	p += a;
	//	ans += a;
	//	if (p >= n) {
	//		ans -= p - n;
	//		cout << ans << endl;
	//		return 0;
	//	}

	//	p += b;
	//	if (p >= n) {
	//		cout << ans << endl;
	//		return 0;
	//	}

	//}

	//1回の操作でa+bだけ列が増える
	//nを超えるための操作数は？

	ll c = n / (a + b);
	if (n % (a + b) != 0) c++;

	//超えた時の列の数は
	ll p = c * (a + b);

	//nよりいくつ大きいか
	ll d = p - n;

	//超えた数が赤（b）より大きければ、青の数も引く
	if (d > b) {
		ans = (c * a) - (d - b);
	}
	else {
		ans = c * a;
	}


	cout << ans << endl;

	return 0;
}