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)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl 	"\n"

typedef vector< ll > 	vi;
typedef vector< vi > 	vvi;
typedef pair< ll,ll > 	ii;

#define pb 		push_back
#define all(c) 		(c).begin(),(c).end()
#define present(c,x) 	((c).find(x) != (c).end())
#define rep(i,n)        for(ll i=0LL; i<(n); i++)

#define MOD 	1000000007
#define INF 	(int)1e9
#define MEM(a, b) memset(a, (b), sizeof(a))

//Maths functions
ll Digit(ll N){ll res=0;while(N) res++, N/=10;return res;}
ll Gcd(ll a,ll b){ return b?Gcd(b,a%b):abs(a); }


int main()
{	 ios_base::sync_with_stdio(false);
	 cin.tie(NULL);
	

	ll n,a,b;
	cin>>n>>a>>b;
	if(a==0)
		cout<<"0";
	else if(b==0)
		cout<<n;
	else if(a+b >= n)
		cout<<a;
	else{
		if(n%(a+b) !=0)
			cout<< ((n+1)/(a+b))*(a);
		else
		cout<<(n/(a+b))*a;
	}
	
	return 0;
}