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>
//----***やべーやつら***----
using namespace std;
#define int long long

//----***型定義***----
using ll = long long;
using P = pair<int,int>;

//----***Like a Pythonista***----
#define REP(ii,jj,nn) for (ll ii=jj;ii<(nn);ii++)
#define RREP(ii,nn,jj) for (ll ii = nn; jj<ii;ii--)
#define each(i,...) for (auto&& i:__VA_ARGS__)
#define ALL(vec) (vec).begin(),(vec).end()
#define sum(...) accumulate(ALL(__VA_ARGS__),0LL)
#define dsum(...) accumulate(ALL(__VA_ARGS__),0.0)
#define vec(type,name,...) vector<type> name(__VA_ARGS__)
template<class T> inline auto max(const T& a){ return *max_element(ALL(a)); }
template<class T> inline auto min(const T& a){ return *min_element(ALL(a)); }
inline ll gcd(ll a,ll b){if(b == 0) return a;return  gcd(b,a%b);}
inline ll lcm(ll a,ll b){ll g = gcd(a,b);return a / g * b;}

//----***定数***----
#define MOD 1000000007
#define INF 100000000000000000
#define EPS 1e-9
const vector<vector<int>> DXY={{-1,0},{1,0},{0,-1},{0,1}};

//----***パーツ***----
#define fi first
#define se second
#define pb push_back
#define re return
#define br break

//----***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...);}
#define debug(var)  do{std::cerr << #var << " ↓ "<<"\n";view(var);}while(0);
#define dbg cerr<<"🥺🥺🥺🥺🥺🥺"<<endl;
template<typename T> void view(T e){std::cout << e << std::endl;}
template<typename T> void view(const std::vector<T>& v){for(const auto& e : v){ std::cout << e << " "; } std::cout << std::endl;}
template<typename T> void view(const std::vector<std::vector<T> >& vv){ for(const auto& v : vv){ view(v); } }

//----***初期時読み込み***----
struct initial{initial(){cin.tie(0); ios::sync_with_stdio(0); cout<<fixed<<setprecision(15);};}initial_;

signed main(){
  int W,H,N;cin>>W>>H>>N;
  vector<vector<int>> G(H,vector<int>(W,1));
  int x,y,a;
  REP(i,0,N){
    cin>>x>>y>>a;
    if(a==1){
      REP(i,0,H){
        REP(j,0,x){
          G[i][j]=0;
        }
      }
    }
    if(a==2){
      REP(i,0,H){
        REP(j,x,W){
          G[i][j]=0;
        }
      }
    }
    if(a==3){
      REP(i,H-y,H){
        REP(j,0,W){
          G[i][j]=0;
        }
      }
    }
    if(a==4){
      REP(i,0,H-y){
        REP(j,0,W){
          G[i][j]=0;
        }
      }
    }
    // debug(G);
  }
  int ans=0;
  REP(h,0,H)REP(w,0,W)ans+=G[h][w];
  print(ans);


}