// Arup Guha
// 1/28/2018
// Solution to Jan 2018 USACO Bronze Problem: Billboard
// This is slower but fewer cases to consider, easy to implement and still runs in time.

import java.util.*;
import java.io.*;

public class billboard_bronze {

	final public static int OFFSET = 1000;
	final public static int MAX = 2001;

	public static void main(String[] args) throws Exception {

		// Read the data.
		Scanner stdin = new Scanner(new File("billboard.in"));

		boolean[] iter = {true, false};
		boolean[][] need = new boolean[MAX][MAX];

		// Go through data.
		for (int i=0; i<2; i++) {

			// Read the coordinates.
			int x1 = stdin.nextInt() + OFFSET;
			int y1 = stdin.nextInt() + OFFSET;
			int x2 = stdin.nextInt() + OFFSET;
			int y2 = stdin.nextInt() + OFFSET;

			// Set these squares to the appropriate value.
			for (int x=x1; x<x2; x++)
				for (int y=y1; y<y2; y++)
					need[x][y] = iter[i];
		}

		boolean zero = true;

		// Find min and max of what needs to be covered.
		int minX = MAX, maxX = -1, minY = MAX, maxY = -1;
		for (int x=0; x<MAX; x++) {
			for (int y=0; y<MAX; y++) {
				if (need[x][y]) {
					zero = false;
					minX = Math.min(minX, x);
					maxX = Math.max(maxX, x);
					minY = Math.min(minY, y);
					maxY = Math.max(maxY, y);
				}
			}
		}

		// Either we don't need to cover anything, or use are mins and maxs.
		int res = zero ? 0 : (maxX-minX+1)*(maxY-minY+1);

		// Ta da!
		PrintWriter out = new PrintWriter(new FileWriter("billboard.out"));
		out.println(res);
		out.close();
		stdin.close();
	}
}