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)

/*
   AUTHOR :- Gaurang_Kothiya
   
   Quote:-" Great things never come from comfort zones."                                                                                                                
*/
 
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>  
#include <ext/pb_ds/tree_policy.hpp> 
#include <functional>
 
using namespace std;
using namespace __gnu_pbds;  

typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> data_set; 
#define speed ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define ll long long  
#define ld long double
#define f(i,x,n) for(ll i=x;i<=n;i++)
#define fe(x,v)   for(auto &x :v)
#define nf(i,x,n) for(ll i=n;i>=x;i--)
#define all(a)   a.begin(),a.end()
#define b_s(a,b)   binary_search( all(a),b)
#define lb(v,n)  lower_bound(all(v) ,n )
#define ub(v,n)  upper_bound( all(v) , n)
#define gcd(a,b) __gcd(a,b)
#define pub push_back
#define pob pop_back
#define puf push_front
#define pof pop_front
#define B begin()
#define E end()
#define acc(v, x) accumulate(all(v), x)
#define p(a,b) pair<a,b>
#define F first
#define S second
#define sz size()
#define v(x) vector<x>
#define m(a,b) map<a,b>
#define u_m(a,b) unordered_map<a,b>
#define mm(a,b) multimap<a,b>
#define s(a) set<a>
#define us(a) unordered_set<a>
#define ms(a) multiset<a>
#define maxy __LONG_LONG_MAX__
#define miny INT_MIN
#define PI 3.14159265358979323846
#define MOD 1000000007



void solver()
{
	ll n,a,b;
	cin>>n>>a>>b;
	
	ll t1= n/(a+b);
	ll t2= n%(a+b);
	ll t3= t1/2 + t1%2 ;
	ll t4= min(a,t2);
	cout<< t3*a + t4;
}

signed main() 
{
	speed	
		
	ll q=1;	
	//cin>>q;	
	
	while(q--)	
	solver();	
	
	//cout << "\n\n"<<"Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " seconds.\n";
	return 0;
}