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 fr first
#define se second
#define r_0 return 0
#define ll long long
#define pb push_back
#define TIME ios_base::sync_with_stdio(0);
using namespace std;

const long long INF = 1e7;
const long long N = 1e6 + 7;
const long long mod = 1e9 + 7;
const long double eps = 1E-7;

ll T = 1, n, m, b, ans, a, cnt;
string s;
vector < ll > v, v1;
set< char > st;
int main(){
    TIME
    cin >> n >> a >> b;
    cnt = a + b;
    ans = n / cnt;
    cout << ans * a + min( n % cnt, a );
}