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>
#include <ext/pb_ds/assoc_container.hpp>
#define INF 1000000
#define watch(x) cout << setw(10) << "[" << (#x) << ":] " << (x) 
#define FORE(i,v,n) for(int i=v;i<n;i++)
#define FORI(i,v,n) for(int i=v;i<=n;i++)
#define vi vector<int>;
#define pii pair<int, int>

using namespace std;
using namespace __gnu_pbds;

//typedef tree <int, null_type, less<int> rb_tree_tag, tree_order_statistics_node_update> indexed_set;

int const MOD = 1000000007;

int test_case(){
	unsigned long long int n,a,b;
	cin >> n >> a >> b;
	if(a == 0) {
		cout << 0 << endl;
		return 0;
	}
	else{
		cout << ((n/(a+b))*a + n%(a+b)); cout << endl;
	}
	
	return 0;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int test=1;
	//cin >> test;
	for(int i=0;i<test;i++){
		test_case();
	}
	return 0;
}