#include <bits/stdc++.h>

using namespace std;

typedef long long lint;
typedef long double llf;
typedef pair<lint, lint> pi;

struct line
{
	long long int a, b;
};

struct cht {
	vector<pi> v;
	int p;
	double cross(int a, int b)
	{
		return (1.0 * (v[a].second - v[b].second)) / (v[b].first - v[a].first);
	}
	lint query(long long int x) {
		if (v.empty()) return 1e18;
		while (p + 1 < v.size() && v[p].first * x + v[p].second > v[p + 1].first * x + v[p + 1].second) p++;
		return v[p].first * x + v[p].second;
	}
	lint query2(long long int x)
	{
		int l = 0;
		int r = v.size() - 1;
		while (l != r)
		{
			int m = (l + r) / 2;
			if (cross(m, m + 1) <= x)
			{
				l = m + 1;
			}
			else
			{
				r = m;
			}
		}
		return v[l].first * x + v[l].second;
	}
	bool chk(pi a, pi b, pi c) {
		return 1ll * (b.first - a.first) * (b.second - c.second) < (c.first - b.first) * (a.second - b.second);
	}
	void add(int x, lint y) {
		if (!v.empty() && v.back().first == x) {
			if (v.back().second > y) return;
			v.pop_back();
		}
		while (v.size() >= p + 2 && chk(v[v.size() - 2], v.back(), pi(x, y))) {
			v.pop_back();
		}
		v.emplace_back(x, y);
	}
}cht;

long long int dp[2][300005];

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

	
	long long int N, A, W, B, t;
	vector <long long int> D;

	cin >> N >> A >> B >> W;
	for (int i = 0; i < N; i++)
	{
		cin >> t;
		D.push_back(t);
	}

	fill(dp[0], dp[0] + 300005, 1e18);
	fill(dp[1], dp[1] + 300005, 1e18);
	dp[1][0] = W;

	/*
	for (int i = 1; i <= N; i++)
	{
		for (int j = 0; j < 2; j++)
		{
			dp[1][i] = min(dp[1][i], dp[j][i - 1] + D[i - 1]);
		}
		for (int j = 0; j < i; j++)
		{
			long long int n = i - j;
			long long int sum = n * (n + 1) / 2;
			sum *= B;
			sum -= (A * n);
			dp[0][i] = min(dp[0][i], dp[1][j] + sum);
		}
	}
	*/

	//dp[0][x] = min(0<=y<x) dp[1][y] + B*(x-y)*(x-y+1)/2 - A*(x-y)
	//dp[1][y] + B/2*(x^2-xy+x-xy+y^2-y) - A*(x-y)
	//dp[1][y] - B*x*y + (B/2)(x^2 + y^2 + x - y) - A * (x-y)
	//dp[1][y] - Bxy + (B/2)(x^2 + x) - Ax + (B/2)(y^2 - y) + A*y
	// A[x] = -Bx , B[y] = y , C[y] = dp[1][y] + (B/2)(y^2 - y) + Ay , D[x] = (B/2)(x^2 + x) - Ax

	line ins;
	ins.a = 0;
	ins.b = dp[1][0];
	cht.add(ins.a, ins.b);
	for (long long int i = 1; i <= N; i++)
	{
		for (int j = 0; j < 2; j++)
		{
			dp[1][i] = min(dp[1][i], dp[j][i - 1] + D[i - 1]);
		}
		//cout << cht.query2(B*i) + (B * ((i * (i + 1)) / 2)) - (A * i) << '\n';
		dp[0][i] = min(dp[0][i], cht.query2(B*i) + (B * ((i * (i + 1)) / 2)) - (A * i));
		line ins;
		ins.a = -i;
		ins.b = dp[1][i] + (B * ((i*(i-1)) / 2)) + A * i;
		cht.add(ins.a, ins.b);
	}

	cout << min(dp[0][N],dp[1][N]) << '\n';

	return 0;
}