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;
typedef long long ll;
typedef vector<ll> vi;
#define REP(i, n) for (ll i = 0; i < (n); ++i)
#define OUT(x) cout << (x) << endl;
#define OUTA(a, n) \
    REP(i, n) { cout << (a[i]) << (i == n - 1 ? "\n" : " "); }
#define ALL(a) (a).begin(), (a).end()
#define SORT(a) sort(ALL(a))
#define RSORT(a) \
    SORT(a);     \
    reverse(ALL(a))

const ll INF = 1e18;
const ll MOD = 1e9 + 7;

int main()
{
    ll N, A, B;
    cin >> N >> A >> B;
    ll ans = (N / (A + B)) * A + min(A, (N % (A + B)));
    OUT(ans);
}