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>
using namespace std;
void Main();

using i8  = int8_t;   /* -128 ~ 127 */
using u8  = uint8_t;  /* 0 ~ 255 */
using i16 = int16_t;  /* -32,768 ~ 32,767 */
using u16 = uint16_t; /* 0 ~ 65,535 */
using i32 = int32_t;  /* -2,147,483,648 ~ 2,147,483,647 */
using u32 = uint32_t; /* 0 ~ 4,294,967,295 */
using i64 = int64_t;  /* -9,223,372,036,854,775,808 ~ 9,223,372,036,854,775,807 */
using u64 = uint64_t; /* 0 ~ 18,446,744,073,709,551,615 */
using f32 = float;    /* (-3.4 * 10^38) ~ (3.4 * 10^38) */
using f64 = double;   /* (-1.7 * 10^308) ~ (1.7 * 10^308) */
using f80 = __float80;

template <class T> using Vec = vector<T>;

/* 定数 */
constexpr i64 INF = 1010000000000000017;
constexpr i64 MOD = 1e9 + 7;
constexpr f64 EPS = 1e-12;
constexpr f64 PI = 3.14159265358979323846;

#define ALL(v) v.begin(), v.end()

#define GCD(a, b) __gcd(a, b) /* 最大公約数 */
#define LCM(a, b) a / __gcd(a, b) * b /* 最小公倍数 */

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(15);
    Main();
    return 0;
}

void Main() {
    u64 N;
    u64 A, B;
    cin >> N >> A >> B;
    u64 l = N / (A + B);
    u64 blue = l * A;
    blue += N % (A + B);
    cout << blue << endl;
}