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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

// inputに入力データ全体が入る
function Main(input) {
  input = input.split("\n");
  N = parseInt(input[0].split(" ")[0]);
  A = parseInt(input[0].split(" ")[1]);
  B = parseInt(input[0].split(" ")[2]);
  count = 0;
  if (N / (A + B) < 1) {
    if(A <= N){
      count = A
    }else{
    count = N
    }
  } else {
    count = A * Math.floor(N / (A + B));
    console.log(N % (A + B))
    if (N % (A + B) <= A) {
        count += N % (A + B);
    }else{
		count += A;
    }
  }

  console.log(count);
}

//*この行以降は編集しないでください（標準入出力から一度に読み込み、Mainを呼び出します）
Main(require("fs").readFileSync("/dev/stdin", "utf8"));