/*input
4 5 3 100
2 80 100
20 125 130
10 75 150
5 120 110
*/
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
namespace my_template {
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;
typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;
#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;
template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) {
return pair<T, T>(a.x + b.x, a.y + b.y);
}
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) {
return pair<T, T>(a.x - b.x, a.y - b.y);
}
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) {
return (a.x * b.x + a.y * b.y);
}
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) {
return (a.x * b.y - a.y * b.x);
}
template<typename T>
void print(vector<T> vec, string name = "") {
cout << name;
for (auto u : vec) {
cout << u << ' ';
}
cout << '\n';
}
}
using namespace my_template;
const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
struct node {
int l, r;
int did;
node *left = nullptr;
node *right = nullptr;
node(int a, int b): l(a), r(b), did(-MOD) {
if (l != r) {
left = new node(l, (l + r) / 2);
right = new node((l + r) / 2 + 1, r);
}
}
void upd(int pos, int val) {
if (l == r) {
assert(l == pos);
did = max(did, val);
return;
}
else if (pos <= (l + r) / 2) {
left->upd(pos, val);
}
else {
right->upd(pos, val);
}
did = max(left->did, right->did);
}
int get(int a, int b) {
if (r < a or b < l) {
return -MOD;
}
else if (a <= l and r <= b) {
return did;
}
else {
return max(left->get(a, b), right->get(a, b));
}
}
};
int N, U, D, S;
int L = 500005;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> N >> U >> D >> S;
vector<pair<int, pi>> sk(N);
for (int i = 0; i < N; ++i)
{
int t, l, m;
cin >> t >> l >> m;
sk[i] = {t, {l, m}};
}
sort(sk.begin(), sk.end(), [](const pair<int, pi> &a, const pair<int, pi> &b) {
return a.x < b.x;
});
node nuoPrad(1, L);
node nuoPab(1, L);
nuoPrad.upd(S, D * S);
nuoPab.upd(S, -U * S);
vpi vec;
for (int h = 0; h < N; ++h)
{
vec.emplace_back(sk[h].y);
if (h + 1 < N and sk[h].x == sk[h + 1].x) {
continue;
}
sort(vec.begin(), vec.end(), [](const pi & a , const pi & b) {
return a.x < b.x;
});
int n = (int)vec.size();
vi pre(n);
vi suf(n);
for (int j = 0; j < n; ++j)
{
int l = vec[j].x;
int m = vec[j].y;
pre[j] = suf[j] = m + max(-l * D + nuoPrad.get(1, l), l * U + nuoPab.get(l, L));
}
for (int i = 1; i < n; ++i)
{
pre[i] = max(pre[i], pre[i - 1] - D * (vec[i].x - vec[i - 1].x) + vec[i].y);
}
for (int i = n - 2; i >= 0; --i)
{
suf[i] = max(suf[i], suf[i + 1] - U * (vec[i + 1].x - vec[i].x) + vec[i].y);
}
for (int i = 0; i < n; ++i)
{
int did = max(pre[i], suf[i]);
nuoPrad.upd(vec[i].x, D * vec[i].x + did);
nuoPab.upd(vec[i].x, -U * vec[i].x + did);
}
vec.clear();
}
int naujasPrad = -S * D + nuoPrad.get(1, S);
int naujasPab = S * U + nuoPab.get(S, L);
printf("%d\n", max(naujasPrad, naujasPab));
}
/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/
