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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
int main(void){
    // Here your code !
    int W,H,N,a[100],x[100],y[100],o[2]={0},X,Y,i;
    scanf("%d%d%d",&W,&H,&N);
    Y = H;
    X = W;
    for(i = 0;i < N;i++){
        scanf("%d%d%d",&x[i],&y[i],&a[i]);
        switch(a[i]){
            case 1:if(o[0] < x[i] && x[i] < X){
                    o[0] = x[i];break;}
                    else{ 
                    printf("0\n");return 0;}
            case 2:if(o[0] < x[i] && x[i] < X){
                    X = x[i];break;}
                    else{ 
                    printf("0\n");return 0;}
            case 3:if(o[1] < y[i] && y[i] < Y){
                    o[1] = y[i];break;}
                    else{ 
                    printf("0\n");return 0;}
            case 4:if(o[1] < y[i] && y[i] < Y){
                    Y = y[i];break;}
                    else{ 
                    printf("0\n");return 0;}
        }
    }
    printf("%d",(X-o[0])*(Y-o[1]));
    return 0;
}