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)

/**
 * Created by hiramekun at 00:27 on 2020-05-19.
 */
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vl = vector<ll>;
using vvl = vector<vl>;
using vb = vector<bool>;
using P = pair<ll, ll>;
template<typename T> using pq = priority_queue<T>;
template<typename T> using minpq = priority_queue<T, vector<T>, greater<T>>;
template<typename T, typename K> using ump = unordered_map<T, K>;
const ll dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const ll mod = 1000000007;
const ll inf = ll(1e9);
const ll e5 = ll(1e5);
const ll ll_inf = ll(1e9) * ll(1e9);

#define rep(i, n) for(ll i = 0; i < (ll)(n); i++)
#define repr(i, n) for(ll i = ll(n - 1); i >= 0; i--)
#define each(i, mp) for(auto& i:mp)
#define eb emplace_back
#define F first
#define S second
#define all(obj) (obj).begin(), (obj).end()

template<class T>
ostream &operator<<(ostream &out, const vector<T> &list) {
    ll n = list.size();
    rep(i, n) out << list[i] << ' ';
    return out;
}

template<class T>
istream &operator>>(istream &in, vector<T> &list) {
    ll n = list.size();
    rep(i, n) in >> list[i];
    return in;
}

template<class T>
ostream &operator<<(ostream &out, const vector<vector<T>> &list) {
    ll n = list.size();
    rep(i, n) out << list[i] << '\n';
    return out;
}

/* ------------- ANSWER ------------- */
/* ---------------------------------- */

void solve() {
    ll n, a, b;
    cin >> n >> a >> b;
    cout << (n / (a + b)) * a + min((n % (a + b)), a) << '\n';
}

int main() {
#ifdef MY_DEBUG
    while (true) {
#endif
        solve();
#ifdef MY_DEBUG
    }
#endif
    return 0;
}