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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define FOR(I,A,B) for(ll I = (A); I < (B); ++I)
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin()) //ai>=v  x is sorted
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin()) //ai>v  x is sorted
#define NUM(x,v) (POSU(x,v)-POSL(x,v))  //x is sorted
#define SORT(x) (sort(x.begin(),x.end())) // 0 2 2 3 4 5 8 9
#define REV(x) (reverse(x.begin(),x.end())) //reverse
#define TO(x,t,f) ((x)?(t):(f))
#define CLR(mat) memset(mat, 0, sizeof(mat))
#define FILV(x,a) fill(x.begin(),x.end(),a)
#define FILA(ar,N,a) fill(ar,ar+N,a)
#define NEXTP(x) next_permutation(x.begin(),x.end())
ll gcd(ll a,ll b){if(a<b)swap(a,b);if(a%b==0)return b;else return gcd(b,a%b);}
ll lcm(ll a,ll b){ll c=gcd(a,b);return ((a/c)*(b/c)*c);}//saisyo kobaisu
#define pb push_back
#define pri(aa) cout<<(aa)<<endl
const ll INF=1e9+7;
const ll N = 100003;
ll w,h,x,y,n,w0=0,h0=0,a;

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>w>>h>>n;

	FOR(i,0,n){
		cin>>x>>y>>a;
		if(a==1){
			if(w0<x)w0=x;
		}else if(a==2){
			if(w>x)w=x;
		}else if(a==3){
			if(h0<y)h0=y;
		}else{
			if(h>y)h=y;
		}
	}

	ll ans=(w-w0)*(h-h0);
	if(ans<0)ans=0;
	pri(ans);
	return 0;
}