Problem:
Score : 100 points

Problem Statement

There are H rows and W columns of white square cells.

You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.

How many white cells will remain?

It can be proved that this count does not depend on what rows and columns are chosen.

Constraints

All values in input are integers.

1 \leq H, W \leq 20

1 \leq h \leq H

1 \leq w \leq W

Input

Input is given from Standard Input in the following format:

H W
h w

Output

Print the number of white cells that will remain.

Sample Input 1

3 2
2 1

Sample Output 1

1

There are 3 rows and 2 columns of cells. When two rows and one column are chosen and painted in black, there is always one white cell that remains.

Sample Input 2

5 5
2 3

Sample Output 2

6

Sample Input 3

2 4
2 4

Sample Output 3

0

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::io::Read;
use std::cmp;

fn main() {
    let mut buf = String::new();

    // 標準入力から全部bufに読み込む
    std::io::stdin().read_to_string(&mut buf);

    // 読み込んだStringを空白で分解する
    let mut iter = buf.split_whitespace();

    // 入力を受け取る
    let large_h: usize = iter.next().unwrap().parse().unwrap();
    let large_w: usize = iter.next().unwrap().parse().unwrap();
    let h: usize = iter.next().unwrap().parse().unwrap();
    let w: usize = iter.next().unwrap().parse().unwrap();

    // マス目の合計
    let sum_mass: usize = large_h * large_w;
    // 黒色の塗りの際に重複が発生するマス目
    let overlap_mass: usize = h * w;
    // 重複を排除した黒色の箇所 = h * large_w + w * large_h - overlap_mass
    let painted_mass: usize = h * large_w + w * large_h - overlap_mass;
    // 残り白色のマス
    let white_mass: usize = sum_mass - painted_mass;

    println!("{}", white_mass);
}