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)

///mAAria...
#include<bits/stdc++.h>
 
using namespace std;
 
#define int long long
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define all(v) v.begin(),v.end()
#define ld long double
#define ff first
#define ss second
#define pii pair<int,int>
#define pi 3.1415926535897932384626433832
#define mod 1000000007
#define inf 1000000000
#define pr1(a) cout << a << endl
#define pr2(a,b)cout << a <<  " " <<b<< endl
#define pr3(a,b,c) cout << a <<" " << b << " " << c << endl

const long double PI = acos(-1);

int powm(int a, int b) {
    int res=1;
    while(b) {
        if(b&1)
            res=(res*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return res;
}

int gcd(int a, int b){
    if(b==0)return a;
    gcd(b,a%b);
}


void solve(){
    int n, b,r;
    cin >> n>> b >> r;

    int ans = 0;

    if(n<=b){
        cout << n << endl;
        return;
    }
    else if(n>b && n<=b+r){
        cout << b << endl;
        return;
    }
    else{

        int x = n/(b+r);
        n%=(b+r);

        ans = x*b + min(n,b);
        cout << ans << endl;
    }


 
} 

signed main() 
{   
    IOS;
    int t = 1;
    //cin >> t;

    for(int i = 1;i<=t;i++){
        solve();
    }
    return 0;
}