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# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoder.Abc
{
    class QuestionB
    {
        public static void Main(string[] args)
        {
            var sw = new System.IO.StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
            Console.SetOut(sw);

            string[] inputStrArray = Console.ReadLine().Split(' ');
            long[] n = new long[inputStrArray.Length];

            for (int i = 0; i < inputStrArray.Length; i++)
            {
                n[i] = long.Parse(inputStrArray[i]);
            }

            string result;

            if (n[1] == 0) result = "0";

            //あまり
            long a = n[0] % (n[1] + n[2]);
            //まとめて計算できる数
            long b = n[0] - a;
            //青いボールの割合
            long c = b / (n[1] + n[2]);
            //青いボールの数
            long d = n[1] * c;


            //あまりの計算
            long
                sumCount = b,
                bCount = 0;

            int k = 0;


            if (n[1] >= a)
            {
                bCount = a;
                goto end;
            }

            while (sumCount < n[0])
            {
                for (k = 0; k < n[1]; k++)
                {
                    bCount++;
                    sumCount++;
                    if (sumCount >= n[0]) goto end;
                }

                for (k = 0; k < n[2]; k++)
                {
                    sumCount++;
                    if (sumCount >= n[0]) goto end;
                }
            }

            end:

            result = (d + bCount).ToString();

            Console.WriteLine(result);

            Console.Out.Flush();
        }
    }
}