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 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
typedef long long ll;
// const int INF = (1 << 30) - 1;
// const ll LL_INF = (1LL << 62) - 1;
// const ll MOD = 1e9 + 7;

ll N, A, B;
// vector<int> A;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> A >> B;
  // A.resize(N);
  // for (int i = 0; i < N; i++) {
  //   int tmp;
  //   cin >> tmp;
  //   A[i] = tmp;
  // }

  ll turn = N / (A + B);
  ll ans = A * turn;
  ans += min(N % (A + B), A);

  cout << ans << endl;
  return 0;
}