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>
#pragma GCC optimize("Ofast")
using namespace std;
mt19937 gen(time(0));
mt19937_64 gen64(228);
long double ans(long long x1, long long x2, long long v) {
  long double len = x2 - x1;
  if (v < len) return 2e9 + 5;
  long double l = max((long double)0, v - len);
  l = min(l, len);
  return l + 2 * (len - l);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(15);
  long long n, k, s, t;
  cin >> n >> k >> s >> t;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i].first >> a[i].second;
  vector<long double> b(k);
  for (int i = 0; i < k; ++i) {
    long long x;
    cin >> x;
    b[i] = x;
  }
  b.insert(b.begin(), 0);
  b.emplace_back(s);
  k += 2;
  sort((b).begin(), (b).end());
  long double l = 0, r = 2e9 + 5;
  for (int z = 0; z < 100; ++z) {
    long double m = (l + r) / 2;
    long double timeTotal = 0;
    for (int i = 0; i + 1 < k; ++i) {
      timeTotal += ans(b[i], b[i + 1], m);
    }
    if (timeTotal <= t) {
      r = m;
    } else {
      l = m;
    }
  }
  long long cost = 2e9 + 5;
  for (auto &elem : a) {
    if (elem.second >= r) {
      cost = min(cost, (long long)elem.first);
    }
  }
  if (cost == 2e9 + 5)
    cout << -1 << '\n';
  else
    cout << cost << '\n';
}
