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 <algorithm>
#include <bitset>
#include <cmath>
#include <complex>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

typedef long long ll;

#define rep(i, a, b) for (ll i = a; i < (ll)b; ++i)
#define INF 10e12
#define MAX 51000
#define all(x) (x).begin(), (x).end()
#define MX(x) *max_element(all(x))
#define MN(x) *min_element(all(x))

using namespace std;

const long long MOD = 1e9 + 7;

int main(void)
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    ll n, a, b;
    cin >> n >> a >> b;
    ll kazu = n / (a + b);
    cout << a * kazu + min(n % (a + b), a) << endl;
}