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>
#define range(i, l, r) for(long long int (i) = (l); (i) < (r); (i)++)
#define reversed_range(i, l, r) for (long long int (i) = (r) - 1; (i) >= l; (i)--)
using namespace std;
template <typename T>
using vec = vector<T>;
using lint = long long;
using ulint = unsigned long long;
using pint = pair<int, int>;
using plint = pair<lint, lint>;

template <typename S, typename T>
ostream& operator <<(ostream& os, pair<S, T> p) {
    os << "(";
    os << p.first << ", " << p.second;
    return os << ")";
}

template <typename T>
ostream& operator <<(ostream& os, vec<T> v) {
    os << "[";
    if (v.size() == 0) return os << "]";
    for (int i = 0; i < v.size() - 1; i++) {
        os << v.at(i) << ", ";
    }
    return os << v.at(v.size() - 1) << "]";
}

template <typename T>
ostream& operator <<(ostream& os, set<T>& s) {
    os << "{";
    if (s.begin() == s.end()) return os << "}";
    auto it_first_item = s.begin();
    os << *it_first_item;
    for (auto it = ++it_first_item; it != s.end(); it++) {
        os << ", " << *it;
    }
    return os << "}";
}

template <typename T>
ostream& operator <<(ostream& os, unordered_set<T>& s) {
    os << "{";
    if (s.begin() == s.end()) return os << "}";
    auto it_first_item = s.begin();
    os << *it_first_item;
    for (auto it = ++it_first_item; it != s.end(); it++) {
        os << ", " << *it;
    }
    return os << "}";
}

template <typename K, typename V>
ostream& operator <<(ostream& os, map<K, V> m) {
    os << "{";
    if (m.begin() == m.end()) return os << "}";
    auto it_first_item = m.begin();
    os << it_first_item->first << ": " << it_first_item->second;
    for (auto it = ++it_first_item; it != m.end(); it++) {
        os << ", " << it->first << ": " << it->second;
    }
    return os << "}";
}

template <typename K, typename V>
ostream& operator <<(ostream& os, unordered_map<K, V> m) {
    os << "{";
    if (m.begin() == m.end()) return os << "}";
    auto it_first_item = m.begin();
    os << it_first_item->first << ": " << it_first_item->second;
    for (auto it = ++it_first_item; it != m.end(); it++) {
        os << ", " << it->first << ": " << it->second;
    }
    return os << "}";
}

lint pow(lint num, lint e, lint MOD) {
    lint res = 1;
    lint cur_num = num;
    while (e) {
        if (e & 1) {
            res *= cur_num;
            res %= MOD;
        }
        cur_num *= cur_num;
        cur_num %= MOD;
        e >>= 1;
    }
    return res;
}

int main() {
    cin.tie(0); cout.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
    lint N, A, B; cin >> N >> A >> B;
    lint num_blocks = N / (A + B);
    lint rest = N % (A + B);
    cout << num_blocks * A + min(A, rest) << "\n";
}