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 <iostream>
#include <string>
#include <cstring>
#include <fstream>
#include <random>
#include <cmath>
#include <iomanip>
#include <climits>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <deque>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <list>
#include <unordered_map>
#include <bitset>
#include <sstream>

typedef  long long ll;
#define mop 1000000007
using namespace std;


int main() {
    
    unsigned long long n,a,b,x,ans;
    cin >> n >> a >> b;
    x = a+b;
    ans = n/x * a;
    ans += min(n - n/x * x, a);
    cout << ans << endl;
}