2 10 1 1 1
1
10
1
1 5 1 8
 4 4 1 0 1
1

1
1 2 1 4
 2 5 1 0 1
2

1
1 4 1 5
 10 10 1 8 4
10
2 3 4 5 6 7 8 9
10
1 1 3 1
2 1 7 1
1 1 9 1
7 1 4 1
10 1 7 1
2 1 7 1
3 1 2 1
5 1 2 1
10 1 5 1
6 1 9 1
 2 4 1 1 1
1
4
1
1 2 1 3
 2 2 0 1 1

1
1
1 2 2 2
 1000 1000 1 1 10
1
2
1
1 900 1 1000
 5 5 1 1 1
3
2
1
1 5 1 1
 2 4 1 1 1
1
2
1
2 3 2 4
 4 4 1 0 1
4

5
1 1 2 2
1 3 2 2
3 3 4 3
3 2 2 2
1 2 2 3

3
 2
 1
 3
4
4
3
3
4
3
3
4
3
 1
 3
 100
 4
 1
 6
4
3
5
4

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using dbl = long double;
using pii = pair<int, int>;
const int N = 100500;
const ll LINF = (ll)1e18;
ll n, m, vel;
int lads, elevs;
ll lad[N], elev[N];
ll query(ll x1, ll y1, ll x2, ll y2) {
  if (x1 == x2) return abs(y1 - y2);
  ll best = LINF;
  int idx0 = lower_bound(lad, lad + lads, y1) - lad;
  for (int idx : {idx0, idx0 - 1}) {
    if (!(0 <= idx && idx < lads)) continue;
    best = min(best, abs(y1 - lad[idx]) + abs(x1 - x2) + abs(lad[idx] - y2));
  }
  idx0 = lower_bound(elev, elev + elevs, y1) - elev;
  for (int idx : {idx0, idx0 - 1}) {
    if (!(0 <= idx && idx < elevs)) continue;
    best = min(best, abs(y1 - elev[idx]) + (abs(x1 - x2) + vel - 1) / vel +
                         abs(elev[idx] - y2));
  }
  return best;
}
void solve() {
  scanf("%lld%lld%d%d%lld", &n, &m, &lads, &elevs, &vel);
  for (int i = 0; i < lads; i++) scanf("%lld", &lad[i]);
  for (int i = 0; i < elevs; i++) scanf("%lld", &elev[i]);
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    ll x1, y1, x2, y2;
    scanf("%lld%lld%lld%lld", &x1, &y1, &x2, &y2);
    printf("%lld\n", query(x1, y1, x2, y2));
  }
}
int main() {
  solve();
  (void)0;
}
