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;
typedef long long int ll;
typedef pair<ll, ll> P;
typedef vector<ll> VI;
typedef vector<VI> VVI;
const ll MOD = 1000000007;
const ll INF = 2147483647;
const ll LINF = 9223372036854775807;
#define REP(i, n) for (int i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()

int main() {
	ll n, a, b; cin >> n >> a >> b;
	ll ans = n/(a+b);
	ans*=a;
	ans+=min(a,n%(a+b));
	cout << ans << endl;
	return 0;
}