#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <list>
#include <set>
#include <map>
#include <numeric>
#include <regex>
#include <tuple>
#include<iomanip>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;
#define MOD 1000000007 // 10^9 + 7
#define INF 1000000000 // 10^9
#define LLINF 1LL<<60

int dp[51][101][101]; // dp[i][j][k] : i番目までのクリスタルを使って(i,j)に行くのにかかる最小費用
					  // いけない場合はINF
int x[51], y[51], c[51];

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int Gx, Gy, N, F; cin >> Gx >> Gy >> N >> F;
	for (int i = 1; i <= N; i++) cin >> x[i] >> y[i] >> c[i];

	for (int i = 0; i <= N; i++) {
		for (int j = 0; j <= Gx; j++) {
			for (int k = 0; k <= Gy; k++) dp[i][j][k] = INF;
		}
	}

	dp[0][0][0] = 0;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j <= Gx; j++) {
			for (int k = 0; k <= Gy; k++) {
				if (dp[i][j][k] != INF) {

					dp[i + 1][j][k] = min(dp[i + 1][j][k], dp[i][j][k]);
					if (j + x[i + 1] <= Gx && k + y[i + 1] <= Gy) {
						dp[i + 1][j + x[i + 1]][k + y[i + 1]] = min(dp[i + 1][j + x[i + 1]][k + y[i + 1]], dp[i][j][k] + c[i + 1]);
					}
					
				}
			}
		}
	}

	ll ans = INF;
	for (int i = 0; i <= Gx; i++) {
		for (int j = 0; j <= Gy; j++) {
			ll tmp = dp[N][i][j] + (abs(Gx - i) + abs(Gy - j))*F;
			ans = min(ans, tmp);
		}
	}

	cout << ans << endl;

	return 0;
}