#include <bits/stdc++.h>
using namespace std;
#define int long long
int inf = 1e18;
int dl[] = {-1,0,1,0};
int dc[] = {0,1,0,-1};
struct stare
{
int i,j,q;
};
int h,w,a,b,c,n;
pair<int,int>p[100005];
int d[505][505][5];
bool om[505][505];
int dmin[505][505];
class Compare
{
public:
bool operator()(pair<int,stare>cine1, pair<int,stare>cine2)
{
return cine1.first < cine2.first;
}
};
void calc_dmin()
{
for (int i = 0; i <= h; i++)
for (int j = 0; j <= w; j++)
dmin[i][j] = inf;
queue<pair<int,int>> qq;
for (int i = 1; i <= n; i++)
{
qq.push(p[i]);
dmin[p[i].first][p[i].second] = 0;
}
while (!qq.empty())
{
pair<int,int> nod = qq.front();
qq.pop();
for (int i = 0; i < 4; i++)
{
int l = nod.first + dl[i],c = nod.second + dc[i];
if (l >= 0 and l <= h and c >= 0 and c <= w and dmin[l][c] == inf)
{
dmin[l][c] = 1 + dmin[nod.first][nod.second];
qq.push({l,c});
}
}
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> h >> w >> a >> b >> c >> n;
for (int i = 1; i <= n; i++)
cin >> p[i].first >> p[i].second,om[p[i].first][p[i].second] = true;
calc_dmin();
for (int i = 0; i <= h; i++)
{
for (int j = 0; j <= w; j++)
{
for (int q = 0; q < 5; q++)
d[i][j][q] = inf;
}
}
priority_queue<pair<int,stare>, vector<pair<int,stare>>, Compare>pq;
stare initial;
initial.i = p[1].first;
initial.j = p[1].second;
initial.q = 0;
pq.push({0,initial});
while (!pq.empty())
{
int timp = -pq.top().first;
stare nod = pq.top().second;
//cout << timp << ' ' << nod.i << ' ' << nod.j << ' ' << nod.q << endl;
pq.pop();
if (d[nod.i][nod.j][nod.q] == inf)
{
d[nod.i][nod.j][nod.q] = timp;
if (nod.q == 0)
{
stare vecin;
vecin.i = nod.i;
vecin.j = nod.j;
for (vecin.q = 1; vecin.q <= 4; vecin.q++)
pq.push({-(timp + b),vecin});
vecin.q = 0;
for (int cv = 0; cv < 4; cv++)
{
vecin.i = nod.i + dl[cv];
vecin.j = nod.j + dc[cv];
if (vecin.i >= 0 and vecin.i <= h and vecin.j >= 0 and vecin.j <= w)
pq.push({-(timp + c),vecin});
}
}
else
{
stare vecin;
vecin.i = nod.i;
vecin.j = nod.j;
vecin.q = 0;
pq.push({-(timp + c * dmin[nod.i][nod.j]),vecin});
vecin.q = nod.q;
vecin.i = nod.i + dl[nod.q - 1];
vecin.j = nod.j + dc[nod.q - 1];
if (vecin.i >= 0 and vecin.i <= h and vecin.j >= 0 and vecin.j <= w)
pq.push({-(timp + a),vecin});
}
}
}
cout << d[p[n].first][p[n].second][0];
/*for (int i = 0; i <= 6; i++)
{
for (int j = 0; j <= 5; j++)
{
for (int q = 0; q <= 4; q++)
cout << i << ' ' << j << ' ' << q << ' ' << d[i][j][q] << endl;
}
}*/
return 0;
}
/**
d[i][j][0/1/2/3/4] = costul minim ca mingea sa fie la pozitia i si fie sa stea la un om (0) fie sa aiba momentum in dir 1/2/3/4
calculez d cu un dijkstra
din starea (i,j,0) ma pot duce in starea (i,j,1/2/3/4) cu cost B
din starea (i,j,dir) ma pot duce in starea (i_vec,j_vec,dir) cu cost A unde (i_vec,j_vec) e vecina pe directia dir cu (i,j)
din starea (i,j,dir) ma pot duce in starea (i,j,0) cu cost 0 doar daca am om in pozitia (i,j)
din starea (i,j,0) ma pot duce in starea (i_vec,j_vec,0) cu cost C unde (i_vec,j_vec) este vecin pe orice directie cu (i,j)
raspunsul meu este d[sN][tN][0]
initial pornesc cu d[s1][t1][0] = 0
**/
