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>
#define FOR(i, m, n) for (int i = m; i < (n); i++)
#define RFOR(i, m, n) for (int i = (m - 1); i >= 0; i--)
#define REP(i, n) FOR(i, 0, n)
#define RREP(i, n) RFOR(i, n, 0)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;

ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }

ll factorial(ll n) {
  ll answer = 1;
  while (n > 1) {
    answer *= n;
    answer %= mod;
    n--;
  }
  return answer;
}

ll combination(ll n, ll r) {
  return factorial(n) / (factorial(r) * factorial(n - r));
}

ll permutation(ll n, ll r) { return factorial(n) / factorial(n - r); }

int main() {
  int w, h, n;
  cin >> w >> h >> n;
  vector<int> xx = {0, w}, yy = {0, h};

  REP(i, n) {
    int x, y, a;
    cin >> x >> y >> a;
    if (a == 1) {
      xx[0] = max(xx[0], x);
    } else if (a == 2) {
      xx[1] = min(xx[1], x);
    } else if (a == 3) {
      yy[0] = max(yy[0], y);
    } else {
      yy[1] = min(yy[1], y);
    }
  }
  int xAbs = xx[1] - xx[0];
  int yAbs = yy[1] - yy[0];
  int anser = max(xAbs * yAbs, 0);
  cout << anser << endl;
  return 0;
}