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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h> //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rec register char
#define rel register ll
#define gc getchar
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
inline int read(){
  int x=0,f=1;char c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return f?x:-x;
  }
inline bool read(ll &x){
  x=0;int f=1;char c=gc();
  for (;!isdigit(c)&&~c;c=gc()) f^=c=='-';
  if (!~c) return 0;
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return x=f?x:-x,1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
  if (!x) return pc(48);
  if (x<0) x=-x,pc('-');
  int bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  }
inline char fuhao(){
  char c=gc();
  for (;isspace(c);c=gc());
  return c;
  }
}using namespace chtholly;
using namespace std;
int mp[999][999],w=read(),h=read();

int main(){
int i,j;
for (int n=read();n--;){
  int x=read(),y=read(),op=read();
  switch(op){
    case 1:{
      for (i=0;i<x;++i){
        for (j=0;j<h;++j){
          mp[i][j]=1; 
		  }
	    }
	  break;
	  }
	case 2:{
	  for (i=x;i<w;++i){
	    for (j=0;j<h;++j){
	      mp[i][j]=1;
		  }
	    }
	  break;
	  }
	case 3:{
	  for (i=0;i<w;++i){
	    for (j=0;j<y;++j){
	      mp[i][j]=1;
		  }
	    }
	  break;
	  }
	case 4:{
	  for (i=0;i<w;++i){
	    for (j=y;j<h;++j){
	      mp[i][j]=1;
		  }
	    }
	  break;
	  }
    }
//  for (i=0;i<w;++i,pl){
//    for (j=0;j<h;++j){
//      write(mp[i][j]),p32; 
//	  }
//    }
  }
int ans=0;
for (i=0;i<w;++i){
  for (j=0;j<h;++j) ans+=!mp[i][j]; 
  }write(ans);
}