Problem:
Score : 200 points

Problem Statement

There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white.

Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i).

Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows:

If a_i = 1, he painted the region satisfying x < x_i within the rectangle.

If a_i = 2, he painted the region satisfying x > x_i within the rectangle.

If a_i = 3, he painted the region satisfying y < y_i within the rectangle.

If a_i = 4, he painted the region satisfying y > y_i within the rectangle.

Find the area of the white region within the rectangle after he finished painting.

Constraints

1 ≦ W, H ≦ 100

1 ≦ N ≦ 100

0 ≦ x_i ≦ W (1 ≦ i ≦ N)

0 ≦ y_i ≦ H (1 ≦ i ≦ N)

W, H (21:32, added), x_i and y_i are integers.

a_i (1 ≦ i ≦ N) is 1, 2, 3 or 4.

Input

The input is given from Standard Input in the following format:

W H N
x_1 y_1 a_1
x_2 y_2 a_2
:
x_N y_N a_N

Output

Print the area of the white region within the rectangle after Snuke finished painting.

Sample Input 1

5 4 2
2 1 1
3 3 4

Sample Output 1

9

The figure below shows the rectangle before Snuke starts painting.

First, as (x_1, y_1) = (2, 1) and a_1 = 1, he paints the region satisfying x < 2 within the rectangle:

Then, as (x_2, y_2) = (3, 3) and a_2 = 4, he paints the region satisfying y > 3 within the rectangle:

Now, the area of the white region within the rectangle is 9.

Sample Input 2

5 4 3
2 1 1
3 3 4
1 4 2

Sample Output 2

0

It is possible that the whole region within the rectangle is painted black.

Sample Input 3

10 10 5
1 6 1
4 1 3
6 9 4
9 4 2
3 1 3

Sample Output 3

64

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int W = Integer.parseInt(sc.next());
		int H = Integer.parseInt(sc.next());
		int N = Integer.parseInt(sc.next());
		int[] X = new int[N];
		int[] Y = new int[N];
		int[] A = new int[N];
		for(int i = 0;i < N;i++) {
			X[i] = Integer.parseInt(sc.next());
			Y[i] = Integer.parseInt(sc.next());
			A[i] = Integer.parseInt(sc.next());
		}
		sc.close();
		int[][] blk = new int[W][H];
		for(int i = 0;i < W;i++) {
			for(int j = 0;j < H;j++) {
				blk[i][j] = 0;
			}
		}
		for(int i = 0;i < N;i++) {
			switch(A[i]){
			case 1:
				for(int j = 0;j < X[i];j++) {
					for(int k = 0;k < H;k++) {
						blk[j][k]++;
					}
				}
				break;
			case 2:
				for(int j = X[i];j < W;j++) {
					for(int k = 0;k < H;k++) {
						blk[j][k]++;
					}
				}
				break;
			case 3:
				for(int j = 0;j < W;j++) {
					for(int k = 0;k < Y[i];k++) {
						blk[j][k]++;
					}
				}
				break;
			case 4:
				for(int j = 0;j < W;j++) {
					for(int k = Y[i];k < H;k++) {
						blk[j][k]++;
					}
				}
				break;
			}
		}
		int ans = 0;
		for(int i = 0;i < W;i++) {
			for(int j = 0;j < H;j++) {
				if(blk[i][j] == 0) {
					ans++;
				}
			}
		}
		System.out.println(ans);
	}
}