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: Wrong Answer)

#include <algorithm>
#include <array>
#include <cmath>
#include <cstdlib>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <queue>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <vector>

#define REP(X, Y, Z) for (ll(X) = (Y); (X) < (Z); (X)++)
#define REV(X, Y, Z) for (ll(X) = (Y); (X) > (Z); (X)--)

using namespace std;
using ll = long long;
using dll = deque<ll>;
using pll = pair<ll, ll>;
using qll = queue<ll>;
using usll = unordered_set<ll>;
using umll = unordered_map<ll, ll>;
using sll = stack<int>;
using vll = vector<ll>;
using vvll = vector<vll>;

constexpr ll kMod = 1e9 + 7;

ll gcd(ll a, ll b) {
  if (a < b) {
    return gcd(b, a);
  }
  ll c = a % b;
  while (c != 0) {
    a = b;
    b = c;
    c = a % b;
  }
  return b;
}

ll lcm(ll a, ll b) {
  return a / gcd(a, b) * b;
}

void init() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}

int main() {
  init();

  ll N, A, B;
  cin >> N >> A >> B;

  if (N < A) {
    cout << N << endl;
    return 0;
  }

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

  ll cnt = N / (A + B);
  ll ans = cnt * A;
  ll idx = 0;
  REP(i, cnt * (A+B), N) {
    if (idx < A) {
      ans++;
    }
  }
  cout << ans << endl;

  return 0;
}