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)


// Problem : B - Count Balls
// Contest : AtCoder Beginner Contest 158
// URL : https://atcoder.jp/contests/abc158/tasks/abc158_b
// Memory Limit : 1024.000000 MB 
// Time Limit : 2000.000000 milisec 
// Powered by CP Editor (https://github.com/coder3101/cp-editor2)

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define REP(i,n) for (ll i = 0; i < n; i++)
#define REPR(i,n) for(ll i = n-1; i >= 0; i--)
#define FOR(i,a,b) for (ll i = a; i < b; i++)
#define FORR(i,a,b) for(ll i = a-1; i>= b; i--)
#define ff first
#define ss second
#define MAXN (ll)1000002
#define MODN (ll)1000000007
#define mp make_pair
#define pb push_back
#define LN(x) (ll)x.length()
#define SZ(x) (ll)x.size()
#define bg(x) (x.begin())
#define rbg(x) (x.rbegin())
#define ENDL cout<<endl
ll n,t=1,tmp;
ll arr[MAXN];
int main() {
   ll n,a,b;
cin>>n>>a>>b;
if(a == 0 && b == 0) {
	cout<<0;
} else {
	ll x = n/(a+b);
	ll ans=x*a;
	x = n - x*(a+b);
	ans+=min(x,a);
	cout<<ans;
}
}