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: Time Limit Exceeded)

#include<stdio.h>
int main(void){
  int x,y,n,i=0,j=0;
  scanf("%d",&x);
  scanf("%d",&y);
  scanf("%d",&n);
  int a[n][3],s1=x,s2=0,t1=0,t2=y;
  while(i<n){
    scanf("%d",&a[i][0]);
    scanf("%d",&a[i][1]);
    scanf("%d",&a[i][2]);
  }
  n=0;
  for(i=0;i<n;i++){
    if(a[n][2]==1&&(s1>(x-a[n][0]))){
      s1=x-a[n][0];
    }
    if(a[n][2]==2&&(t1<a[n][0])){
      t1=a[n][0];
    }
    if(a[n][2]==3&&(s2<a[n][1])){
      s2=a[n][0];
    }
    if(a[n][2]==4&&(t2>y-a[n][1])){
      t2=y-a[n][1];
    }      
  }
  int tate=s1-t1,yoko=t2-s2;
  if(tate>0&&tate>0){
    printf("%d",tate*yoko);
  }else{
    printf("0");
  }
  return 0;
}