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;
long long power(long long x, long long y, long long p);
int Bit(long long x);
long long Sum(int x);
inline int D() {
  int t;
  scanf("%d", &t);
  return t;
}
inline long long llD() {
  long long t;
  scanf("%lld", &t);
  return t;
}
const int N = 1e6 + 5;
int arr[N];
vector<pair<long long, long long>> vec;
int n, k, l, r, mid, ans;
long long s, t, a, two, one, cap, MN = 1e9 + 10;
bool flage;
bool can(int idx) {
  cap = vec[idx].first, two = one = 0, flage = true;
  for (int i = 1; i <= k; ++i) {
    flage &= (arr[i] - arr[i - 1] <= cap);
    if (cap >= 2ll * (arr[i] - arr[i - 1])) {
      two += arr[i] - arr[i - 1];
      continue;
    }
    a = cap - (arr[i] - arr[i - 1]);
    two += a;
    one += cap - 2 * a;
  }
  if (flage && two + 2 * one <= t) return true;
  return false;
}
int BS() {
  l = 0, r = n - 1, mid;
  while (l < r) {
    mid = (l + r) >> 1;
    if (can(mid))
      r = mid;
    else
      l = mid + 1;
  }
  if (can(r)) return r;
  return -1;
}
int main() {
  n = D(), k = D(), s = llD(), t = llD();
  for (int i = 0; i < n && scanf("%lld%lld", &two, &one); ++i)
    vec.push_back({one, two});
  for (int i = 1; i <= k; ++i) scanf("%d", arr + i);
  sort(vec.begin(), vec.end());
  sort(arr + 1, arr + k + 1);
  arr[++k] = s;
  ans = BS();
  if (ans != -1)
    while (ans < n) MN = min(MN, vec[ans++].second);
  else
    MN = -1;
  printf("%lld", MN);
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int Bit(long long x) {
  if (!x) return 0;
  return 1 + Bit(x >> 1);
}
long long Sum(int x) { return (1ll * x * (x + 1)) >> 1ll; }
