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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <algorithm>
#include <utility> 
#include <unordered_set>
#include <vector>
#include <list> 
#include <string>
#include <iterator> 
#include <deque>
#include <array>
#include <forward_list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cmath>
#include <climits>
#include <bitset>
#include <numeric>
using namespace std;


const long double PI = 3.14159265358979323846264338327950;
const int N = 2 * 1e5;


template <class T>
T gcd(T a, T b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}


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

	 long long n, a, b;
	 long long ans;
	cin >> n >> a >> b;
	ans = 0;
	if (b == 0 || a == 0) {
		if (a > 0)
			ans = n;
	} else {
		ans = a * (n / (a + b));
		n = (n - 1) % (a + b);

		cout << "n " << n << '\n';
		if (n < a)
			ans += n + 1;
	}

	cout << ans << '\n';
}