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 pp pair<int,int>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define ll long long
#define ld long double
#define all(a) (a).begin(),(a).end()
#define mk make_pair
ll MOD=1000000007;
ll mod=998244353;
int inf=1000001000;
ll INF=1e18+5;

int main() {
  ll n,a,b;
  cin >> n >> a >> b;
  ll ans=(n/(a+b))*a+min(a,n%(a+b));
  cout << ans << endl;
}