C++ :

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <deque>

using namespace std;

const int MAXN = 200000 + 1000;
const int INF = 0x3f3f3f3f;

int n;
int cost1, time1, cost2, time2, pri;
int s[MAXN];

deque<pair<int, int> > que3, que2, que1;

int calc(int x)
{
	int ret = 0;
	que1.clear(), que2.clear(), que3.clear();
	for(int i = 1; i <= n; i++)
	{
		while(!que3.empty() && que3.front().first <= i - time2)
			que2.push_back(que3.front()), que3.pop_front();
		while(!que2.empty() && que2.front().first <= i - time1)
			que1.push_back(que2.front()), que2.pop_front();
		int rem = s[i];
		if(x > 0)
		{
			int del = min(rem, x);
			rem -= del, x -= del;
			ret += del * pri;
		}
		while(rem > 0 && !que1.empty())
		{
			int del = min(rem, que1.back().second);
			rem -= del, que1.back().second -= del;
			ret += del * cost1;
			if(!que1.back().second)
				que1.pop_back();
		}
		while(rem > 0 && !que2.empty())
		{
			int del = min(rem, que2.back().second);
			rem -= del, que2.back().second -= del;
			ret += del * cost2;
			if(!que2.back().second)
				que2.pop_back();
		}
		if(rem)
			return INF;
		que3.push_back(make_pair(i, s[i]));
	}
	return ret;
}

int ternary(int l, int r)
{
	int ret = INF;
	while(l <= r)
	{
		int mid1 = (2 * l + r) / 3, mid2 = (l + 2 * r + 1) / 3;
		int val1 = calc(mid1), val2 = calc(mid2);
		if(val1 > val2 || val1 == INF)
			l = mid1 + 1, ret = min(ret, val2);
		else
			r = mid2 - 1, ret = min(ret, val1);
	}
	return ret;
}

int main()
{
	ios::sync_with_stdio(false);
//	 freopen("napkin.in", "r", stdin);
//	 freopen("napkin.out", "w", stdout);

	cin >> n >> time1 >> time2 >> cost1 >> cost2 >> pri;

	if(time1 >= time2 && cost1 >= cost2)
		time1 = INF;
	else if(time1 <= time2 && cost1 <= cost2)
		time2 = time1, cost2 = cost1, time1 = INF;
	else if(time1 < time2)
		swap(time1, time2), swap(cost1, cost2);

	for(int i = 1; i <= n; i++)
		cin >> s[i];

	int s_sum = 0;
	for(int i = 1; i <= n; i++)
		s_sum += s[i];

	int ans = ternary(1, s_sum);
	cout << ans << endl;


	return 0;
}