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)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define pb push_back
#define all(v) v.begin(), v.end()
#define bigger (char)toupper
#define smaller (char)tolower
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
typedef unsigned long long ull;

int main() {
    int N,A,B;
    scanf("%d%d%d",&N,&A,&B);
    if(A>=N) cout<<N<<endl;
    else if(A+B>=N) cout<<A<<endl;
    else {
        int answer=0;
        int N2=N;
        N2/=(A+B);
        answer+=(N2*A);
        N-=(A+B)*N2;
        answer+=min(N,A);
        cout<<answer<<endl;
    }
}