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)

/*
 * @Author: Kaizyn
 * @Date: 2020-03-07 21:08:03
 * @LastEditors: Kaizyn
 * @LastEditTime: 2020-03-07 21:09:48
 * @FilePath: \Code\atcoder\abc158B.cpp
 */
#include <bits/stdc++.h>

// #define DEBUG

using namespace std;

const int N = 1e2;
const int MOD = 1e9+7;
const double eps = 1e-7;
const int INF = 0x3f3f3f3f;
typedef pair<int, int> pii;

long long n, a, b;

inline void solve()
{
    cin >> n >> a >> b;
    long long t = n/(a+b), r = n-t*(a+b);
    cout << t*a+min(a, r) << endl;
}

signed main()
{
    ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    int testcase = 1;
    // cin >> testcase;
    for (int i = 1; i <= testcase; ++i) {
        solve();
    }
    return 0;
}