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)

/*
	Name:
	Author: xiaruize
	Date:
*/

#include<bits/stdc++.h>
using namespace std;
#define ull unsigned long long
#define MOD 1000000007
#define ALL(a) (a).begin(), (a).end()
#define forn(i, n) for (int i = 0; i < int(n); i++)
#define ll long long
#define form(i,j,n) for(int i=int(j);i<=int(n);i++)
#define pb push_back
#define mk make_pair
#define pii pair<int,int>
#define pis pair<int,string>
#define sec second

inline int read()
{
    char ch=getchar();
    int x=0,f=1;
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+(ch^48),ch=getchar();
    return x*f;
}

int main()
{
 	ios::sync_with_stdio(false);
	cin.tie(0);
 	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	ll n,a,b;
	cin>>n>>a>>b;
	int ans=0;
	int m=n%(a+b);
	if(m>=a)
	{
		ans=a;
	}
	else
	{
		ans=m;
	}
	cout<<n/(a+b)*a+ans<<endl;
	return 0;
}