1 1 2 2
1000000000 1000000000
1
 1 1 1000000000 1000000000
100 1000000000
1
 1 1 10 18
5 6
5
 2 1 1000000000 2000000000
111 999999999
101 999999998
1
 4 13 400 600
13 30
1 19
1 160
1 113
58 73 15 319 194 362 128 157 336 162 77 90 96
 1 15 100 200
283 8
30 58 16 45 80 82 55 95 24 23 86 28 51 47 20
 3 2 300 400
24 68
13 65
15 113
127 177
 1 1 2 1
1 10
1
 4 2 7 15
10 9
4 4
9 3
4 10
1 6
 2 1 1000000000 2000000000
111 1000000000
101 1000000000
5
 2 1 1000000000 2000000000
111 999999998
101 999999998
1
 1 1 1000000000 1000000000
1 1000000000
1

1000000000
 -1
 5
 111
 1
 -1
 -1
 -1
 4
 101
 -1
 -1

#include <bits/stdc++.h>
using namespace std;
pair<int, int> p[200001];
int x[200001];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n, k, s, t;
  cin >> n >> k >> s >> t;
  for (int i = 0; i < n; i++) {
    cin >> p[i].first >> p[i].second;
  }
  vector<pair<int, int> > vt;
  for (int i = 0; i < n; i++) {
    if (vt.empty() || p[i].second > vt.back().second) {
      vt.push_back(p[i]);
    }
  }
  sort(p, p + n);
  for (int i = 1; i <= k; i++) {
    cin >> x[i];
  }
  sort(x + 1, x + k + 1);
  x[k + 1] = s;
  int l = -1;
  int r = vt.size();
  while (l < r - 1) {
    int m = (l + r) / 2;
    int ct = 0;
    for (int j = 0; j <= k; j++) {
      if (vt[m].second < x[j + 1] - x[j]) {
        ct = t + 1;
        break;
      }
      int dt = min(vt[m].second - (x[j + 1] - x[j]), (x[j + 1] - x[j]));
      ct += 2 * (x[j + 1] - x[j]) - dt;
    }
    if (ct > t) {
      l = m;
    } else {
      r = m;
    }
  }
  if (r == vt.size())
    cout << "-1";
  else
    cout << vt[r].first;
  return 0;
}
