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 <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <set>
#include <vector>
#define REP(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;

// 便利関数
template <class T> inline bool chmin(T &a, T b) {if (a > b){a = b;return true;}return false;}
template <class T> inline bool chmax(T &a, T b) {if (a < b){a = b;return true;}return false;}

// 入出力
void print() { std::cout << '\n'; }
template <class T>void print(const T &x) {std::cout << x <<'\n';}
template <class T, class... Args>void print(const T &x, const Args &... args) {std::cout << x << " ";print(args...);}

void solve(long long W, long long H, long long N, std::vector<long long> x, std::vector<long long> y, std::vector<long long> a){
    ll minX = 0, maxX = W, minY = 0, maxY = H;
    REP(i, N) {
        if (a[i] == 1) chmax(minX, x[i]);
        if (a[i] == 2) chmin(maxX, x[i]);
        if (a[i] == 3) chmax(minY, y[i]);
        if (a[i] == 4) chmin(maxY, y[i]);
    }
    if (minX >= maxX || minY >= maxY) print(0);
    else {
        ll ans = (maxX - minX) * (maxY - minY);
        print(ans);
    }
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    long long W;
    scanf("%lld",&W);
    long long H;
    scanf("%lld",&H);
    long long N;
    scanf("%lld",&N);
    std::vector<long long> x(N);
    std::vector<long long> y(N);
    std::vector<long long> a(N);
    for(int i = 0 ; i < N ; i++){
        scanf("%lld",&x[i]);
        scanf("%lld",&y[i]);
        scanf("%lld",&a[i]);
    }
    solve(W, H, N, std::move(x), std::move(y), std::move(a));
    return 0;
}