#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
#define int long long
#define nl '\n'
#define io ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
const int mod = 1000000007, mod2 = 998244353;
// change this
const int N = 3005;
int n, m, k, a, b, c, t, ans;
set< pair<int, int> > s;
signed main() {
io;
cin >> n >> m >> k;
cin >> a >> b >> c;
k -= m;
cin >> t;
for (int i=0; i<m; i++) {
int temp;
cin >> temp;
s.insert({temp-1, 0});
}
// first, let's get current number without adding more
for (auto it = (++s.begin()); it != s.end(); ++it) {
// how many can we get from [lb, ub)?
int lb = (*(--it)).first; it++;
int ub = (*it).first;
int d = min((t - b*lb) / a, (ub-lb-1)); // maximum d such that we can go to lb + d
if (lb * b + d * c > t) continue;
if (d >= 0) ans += (d+1);
}
if (b * (*(--s.end())).first <= t) ans++; // can reach very end also
// cout << "INITIAL " << ans << nl;
// now, for each k, find best.
while (k--) {
pair<int, int> bestin = {-1, -1};
int best = - 1;
for (auto it = (++s.begin()); it != s.end(); ++it) {
// how many can we gain?
auto lb = *(--it); it++;
auto ub = *it;
if (ub.second) continue; // otherwise we would've added this alr
// first, where should we add?
int ni = lb.second + ((t - b*lb.first - c*lb.second) / a) + 1;
if (lb.first * b + ni * c > t) continue; // we can't come here anyway
if (lb.first + ni >= ub.first) continue; // saturated liao
// we should add at ni. how many new stuff does that give us?
int d = (t - b*lb.first - c*ni) / a + 1;
d = min(d, ub.first - (lb.first + ni));
if (d <= 0) continue;
// it gives us d more things!
if (best < d) {
best = d;
bestin = {lb.first, ni};
}
}
if (best == -1) break;
ans += best;
s.insert(bestin);
// cout << bestin.first << ' ' << bestin.second << ' ' << best << nl;
}
cout << ans-1 << nl;
}
