Problem:
Score : 100 points

Problem Statement

The window of Takahashi's room has a width of A. There are two curtains hung over the window, each of which has a horizontal length of B. (Vertically, the curtains are long enough to cover the whole window.)

We will close the window so as to minimize the total horizontal length of the uncovered part of the window.
Find the total horizontal length of the uncovered parts of the window then.

Constraints

1 \leq A \leq 100

1 \leq B \leq 100

A and B are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the total horizontal length of the uncovered parts of the window.

Sample Input 1

12 4

Sample Output 1

4

We have a window with a horizontal length of 12, and two curtains, each of length 4, that cover both ends of the window, for example. The uncovered part has a horizontal length of 4.

Sample Input 2

20 15

Sample Output 2

0

If the window is completely covered, print 0.

Sample Input 3

20 30

Sample Output 3

0

Each curtain may be longer than the window.

Solution in C (Clang 3.8.0) (judge verdict: Runtime Error)

#include <stdio.h>

int main(void){
  int N = 0; // たこ焼きの個数
  int d[50]; // おいしさの値を格納する配列
  int heal_points_list[1225]; // 回復量を格納する配列
  int sum_heal_points = 0; // 回復量の総和

  // たこ焼きの個数を設定する
  scanf("%d", &N);

  // たこ焼き達においしさ値を設定する
  for(int i = 0;i < N; i++) {
    scanf("%d", &d[i]);
  }

  // 総当たり方式の実行回数を求める
  int calc_count = (N * (N - 1)) / 2;

  // 各たこ焼き同士の回復量を総当たり方式で計算する
  for(int i = 0; i < N; i++) {
    int num = i + 1;

    // i が配列の要素数と同じかそれ以上になった場合、numに ０を代入する
    if(calc_count <=  i + 1) num = 0;

    heal_points_list[i] = d[i] * d[num];
  }

  // 回復量の総和を計算する
  for(int i = 0; i < calc_count; i++) {
    sum_heal_points += heal_points_list[i];
  }

  // 出力
  printf("%d\n", sum_heal_points);

  return 0;
}