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>
using namespace std;
#define int long long int
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define r0 return 0;
#define f(k,i,n,j) for(ll k=i;k<=n;k=k+j)
#define cin(a) cin>>a;
#define loop while(t--)
#define vi vector<int>
#define vii vector<long long int>
#define SORT(v) sort(v.begin(),v.end());
#define mod (1e9+7)
#define crap ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define endl "\n"
signed main()
{
    crap;
    int n,a,b;
    cin>>n>>a>>b;
    int ans=n/(a+b);
    ans*=a;
    ans+=min(a,n%(a+b));
    cout<<ans<<endl;
}