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: Accepted)

#include <stdio.h>

int main(){
  int W,H,N;
  scanf("%d %d %d",&W,&H,&N);

  int x[N],y[N],a[N];
  for(int i=0; i<N; i++){
    scanf("%d %d %d",&x[i],&y[i],&a[i]);
  }

  int color[W][H];
  for(int i=0; i<W; i++){
    for(int j=0; j<H; j++){
      color[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++){
	  color[j][k] = 1;
	}
      }
      break;
      
    case 2:
      for(int j=x[i]; j<W; j++){
	for(int k=0; k<H; k++){
          color[j][k] = 1;
	}
      }
      break;

    case 3:
      for(int j=0; j<W; j++){
	for(int k=0; k<y[i]; k++){
          color[j][k] = 1;
	}
      }
      break;

    default:
      for(int j=0; j<W; j++){
	for(int k=y[i]; k<H; k++){
          color[j][k] = 1;
	}
      }
      break;
    }

    /*for(int j=0; j<H; j++){
      for(int i=0; i<W; i++){
	printf("%d ",color[i][j]);
      }
      printf("\n");
    }
    printf("\n");*/

  }

  int count = 0;
  for(int i=0; i<W; i++){
    for(int j=0; j<H; j++){
      if(!color[i][j]){count++;}
    }
  }
  printf("%d\n",count);

  return 0;
}