#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
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<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 4e18;
const int MX = 200005;
bool Q;
struct Line {
mutable ll k, m, p; // slope, y-intercept, last optimal x
bool operator<(const Line& o) const {
return Q ? p < o.p : k < o.k;
}
};
struct LineContainer : multiset<Line> {
const ll inf = LLONG_MAX;
ll div(ll a, ll b) { // floored division
if (b < 0) a *= -1, b *= -1;
if (a >= 0) return a/b;
return -((-a+b-1)/b);
}
// updates x->p, determines if y is unneeded
bool isect(iterator x, iterator y) {
if (y == end()) { x->p = inf; return 0; }
if (x->k == y->k) x->p = x->m > y->m ? inf : -inf;
else x->p = div(y->m - x->m, x->k - y->k);
return x->p >= y->p;
}
void add(ll k, ll m) {
k *= -1, m *= -1;
auto z = insert({k, m, 0}), y = z++, x = y;
while (isect(y, z)) z = erase(z);
if (x != begin() && isect(--x, y)) isect(x, y = erase(y));
while ((y = x) != begin() && (--x)->p >= y->p) isect(x, erase(y));
}
ll query(ll x) { // gives max value
if (empty()) return INF;
Q = 1; auto l = *lb({0,0,x}); Q = 0;
return -(l.k * x + l.m);
}
};
LineContainer L;
ll X,N,M,W,T, cum[MX], dp[MX];
vpl D, S;
void mn(ll& a, ll b) { a = min(a,b); }
void input() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> X >> N >> M >> W >> T;
S.resize(N+1); F0R(i,N) cin >> S[i].s;
S[N].s = X;
F0R(i,N+1) S[i].f = S[i].s % T;
D.resize(M+1); F0R(i,M) cin >> D[i].f >> D[i].s;
D[M] = {T,0};
sort(all(S)), sort(all(D));
F0R(i,M+1) {
if (i) cum[i] = cum[i-1];
cum[i] += D[i].s;
}
}
int main() {
input();
int ind = N+1;
dp[M] = (X/T+1)*W;
F0Rd(i,M) {
ll val = min(L.query(i)-cum[i],dp[i+1]);
ll bes = INF;
while (ind && S[ind-1].f >= D[i].f) mn(bes,S[--ind].s/T*W);
if (bes != INF) L.add(-bes,val+bes*i+cum[i]);
dp[i] = val+((X-D[i].f)/T+1)*W;
}
ll val = min(L.query(-1),dp[0]);
cout << val;
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
* if you have no idea just guess the appropriate well-known algo instead of doing nothing :/
*/
