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;
const int N = 100010;
long long arr1[N], arr2[N];
int main() {
  long long n, m, c1, c2, v;
  int i, j, l, q;
  long long x1, x2, y1, y2;
  scanf("%lld%lld%lld%lld%lld", &n, &m, &c1, &c2, &v);
  for (i = 0; i < c1; i++) scanf("%d", &arr1[i]);
  for (i = 0; i < c2; i++) scanf("%d", &arr2[i]);
  scanf("%d", &q);
  while (q--) {
    long long ans = 10000000;
    scanf("%lld%lld%lld%lld", &x1, &y1, &x2, &y2);
    if (x1 == x2) {
      printf("%lld", abs(y2 - y1));
      continue;
    }
    long long k = lower_bound(arr1, arr1 + c1, y1) - arr1;
    if (k < c1)
      ans = min(ans, abs(arr1[k] - y1) + abs(arr1[k] - y2) + (abs(x1 - x2)));
    if (k > 0) {
      k--;
      ans = min(ans, abs(arr1[k] - y1) + abs(arr1[k] - y2) + (abs(x1 - x2)));
    }
    k = lower_bound(arr2, arr2 + c2, y1) - arr2;
    if (k < c2)
      ans = min(ans, abs(arr2[k] - y1) + abs(arr2[k] - y2) +
                         (abs(x1 - x2) + v - 1) / v);
    if (k > 0) {
      k--;
      ans = min(ans, abs(arr2[k] - y1) + abs(arr2[k] - y2) +
                         (abs(x1 - x2) + v - 1) / v);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
