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>
#define rep(i,n) for (int i = 0; i < (n); i++)
#define repf(i,a,b) for (int i = (a); i < (b); i++)
#define repi(i,a,b) for (int i = (a); i <= (b); i++)
#define repr(i,a,b) for (int i = (a); i >= (b); i--)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define debug(x)  cout << #x << " = " << (x) << endl;
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using pii = pair<int, int>;
using vc = vector<char>;
using vvc = vector<vc>;
using vs = vector<string>;
using tiii = tuple<int, int, int>;

int main() {
  int W, H, N; cin >> W >> H >> N;
  vi x(N), y(N), a(N); rep(i,N) cin >> x.at(i) >> y.at(i) >> a.at(i);
  
  int xs = 0, ys = 0;
  int xe = W, ye = H;
  
  rep(i,N) {
    if (a.at(i) == 1) 
      xs = max(xs, x.at(i));
    else if (a.at(i) == 2)
      xe = min(xe, x.at(i));
    else if (a.at(i) == 3)
      ys = max(ys, y.at(i));
    else
      ye = min(ye, y.at(i));
  }
  
  cout << max(xe-xs,0)*max(ye-ys,0) << endl;
}