Score : $100$ points

问题陈述

给定一个 $2 \times 2$ 矩阵 $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$。 矩阵 $A$ 的行列式可以计算为 $ad-bc$。 求出它。

以上为大语言模型 kimi 翻译，仅供参考。

Problem Statement

Given is a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
The determinant of $A$ can be found as $ad-bc$.
Find it.

Constraints

• All values in input are integers.
• $-100 \le a, b, c, d \le 100$

Input

Input is given from Standard Input in the following format:

$a$ $b$

$c$ $d$

Output

Print the answer as an integer.

Sample Input 1

1 2
3 4

Sample Output 1

-2

The determinant of $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ is $1 \times 4 - 2 \times 3 = -2$.

Sample Input 2

0 -1
1 0

Sample Output 2

1

Sample Input 3

100 100
100 100

Sample Output 3

0