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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i, n) for (ll i = 0; i < (ll)(n); ++i)
#define rep2(i, s, n) for (ll i = s; i < (ll)(n); i++)
#define repr(i, n) for (ll i = n; i >= 0; i--)
#define pb push_back
#define COUT(x) cout << (x) << endl
#define COUTF(x) cout << setprecision(15) << (x) << endl
#define ENDL cout << endl
#define DF(x) x.erase(x.begin())  // 先頭文字削除
#define ALL(x) x.begin(), x.end()
#define SORT(x) sort(ALL(x))
#define REVERSE(x) reverse(ALL(x))
#define ANS cout << ans << endl
#define RETURN(x)    \
  cout << x << endl; \
  return 0
#define init() \
  cin.tie(0);  \
  ios::sync_with_stdio(false)
#define debug(x) cerr << "[debug] " << #x << ": " << x << endl;
#define debugV(v)                       \
  cerr << "[debugV] " << #v << ":";     \
  rep(i, v.size()) cerr << " " << v[i]; \
  cerr << endl;
using namespace std;
using ll = long long;
using ld = long double;
using vll = vector<ll>;
using P = pair<ll, ll>;
constexpr ll INF = 0x3f3f3f3f3f3f3f3f;
constexpr ld PI = 3.141592653589793238462643383279;
ll get_digit(ll x) {
  return to_string(x).size();
}

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

ll lcm(ll a, ll b) {
  return a / gcd(a, b) * b;
}

vector<P> factorize(ll n) {
  vector<P> result;
  for (ll i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      result.pb({i, 0});
      while (n % i == 0) {
        n /= i;
        result.back().second++;
      }
    }
  }
  if (n != 1) {
    result.pb({n, 1});
  }
  return result;
}

vll divisor(ll n) {
  vll ret;
  for (ll i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n) ret.push_back(n / i);
    }
  }
  SORT(ret);
  return (ret);
}

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;
}

signed main() {
  init();
  ll W, H, N;
  cin >> W >> H >> N;

  ll a1 = 0;
  ll a2 = W;
  ll a3 = 0;
  ll a4 = H;

  rep(i, N) {
    ll x, y, a;
    cin >> x >> y >> a;
    if (a == 1) a1 = max(a1, x);
    if (a == 2) a2 = min(a2, x);
    if (a == 3) a3 = max(a3, y);
    if (a == 4) a4 = min(a4, y);
  }
  W = W - a1 - (W - a2);
  H = H - a3 - (H - a4);
  if (0 < H && 0 < W) {
    COUT(H * W);
  } else {
    COUT(0);
  }

  return 0;
}