#include <bits/stdc++.h>
#define endl "\n"
#define ll long long
using namespace std;
struct pass{
ll d,c;
bool operator<(const pass p)const{
return d < p.d;
}
};
struct Line{
ll m,p;
};
ll len = 1; // lenght of leechaotree
struct leechaotree{
vector<Line> tree;
leechaotree(ll l){
while(len < l){
len *= 2;
}
tree = vector<Line>(2*len,{0,(ll)1e18});
}
ll calc(Line line,ll x){// calc mx+p
return 1LL*(-x+1)*line.m+line.p;
}
void insert(Line nw,ll l=0,ll r=len-1,int v=1){
ll mid = l+(r-l)/2;
if(calc(nw,mid) < calc(tree[v],mid)){ // nw is better than tree[v]
swap(tree[v],nw); // remplace tree[v] and continue with it
}
if(r==l) // you're in a leaf
return;
else if(calc(nw,l) < calc(tree[v],l)){ // intersection at left
insert(nw,l,mid,v*2);
}
else // intersection of nw and tree[v] on the right or non-existent
insert(nw,mid+1,r,v*2+1);
}
ll query(ll x,ll l=0,ll r=len-1,int v=1){ // find best on the path
if(l==r) // at a leave
return (-x+1)*tree[v].m+tree[v].p;
ll mid = l+(r-l)/2;
if(x <= mid) // x is on the left
return min((-x+1)*tree[v].m+tree[v].p,query(x,l,mid,v*2));
else // x is on the right
return min((-x+1)*tree[v].m+tree[v].p,query(x,mid+1,r,v*2+1));
}
};
void solve(){
ll X,N,M,W,T;
cin >> X >> N >> M >> W >> T;
vector<ll> S(N);
for(int i=0;i<N;i++)
cin >> S[i];
S.push_back(X);
vector<pass> passager(M);
for(int i=0;i<M;i++){
cin >> passager[i].d >> passager[i].c;
}
sort(passager.begin(),passager.end());
sort(S.begin(),S.end(),[&](const ll& a,const ll& b){ return a%T < b%T;});
vector<ll> pref(M+1,0);
for(int i=1;i<=M;i++){
pref[i] = pref[i-1]+passager[i-1].c;
}
vector<ll> last(N+1,0); // represent the most left passenger before stop i
for(int i=0;i<=N;i++){
pass p = {S[i]%T,-1LL};
last[i] = upper_bound(passager.begin(),passager.end(),p)-passager.begin()-1;
}
vector<ll> dp(M+1,1e18);
dp[M] = 0;
leechaotree lct(M+1);
for(int i=M-1,j=N;i>=0;i--){ // i = passengers, j = stops
while(j >= 0 && i <= last[j]){ // while there's a stop at left of i
ll Dlst = passager[last[j]].d;
//line representation :
//
// dp[i] = pref[last[j]+1]-pref[i] + ((last[j]-i+1)*((S[j]-Dlst)/T)*W)
// + dp[i-1]
// = ( (-i+1) * ((S[j]-Dlst)/T)*W
// + ((S[j]-Dlst)/T)*W+dp[last[j]+1]+pref[last[j]+1] ) - pref[i]
//
// dp[i] = m*x+p - pref[i]
// x = (-i+1)
// m = ((S[j]-Dlst)/T)*W
// p = last[j]*((S[j]-Dlst)/T)*W + dp[last[j]+1]+pref[last[j]+1]
Line line = {((S[j]-Dlst)/T)*W,last[j]*((S[j]-Dlst)/T)*W
+dp[last[j]+1]+pref[last[j]+1]};
lct.insert(line);
j--;
}
ll Di = passager[i].d;
dp[i] = dp[i+1]+((X-Di)/T+1)*W; // passenger[i] don't leave
dp[i] = min(dp[i],-pref[i]+lct.query(i)); // passenger[i] leave at perfect
// time or don't
}
cout << dp[0] + ((X/T+1) * W) << endl; // best cost for passengers + driver cost
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
int t = 1;
//cin >> t;
while(t--){
solve();
}
return 0;
}
