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;
struct ca {
  int c;
  int v;
} car[200005];
int gas[200005];
int Time(int j, int vv) {
  int ss = gas[j] - gas[j - 1];
  if (vv >= 2 * ss)
    return ss;
  else if (ss <= vv && vv < 2 * ss)
    return 3 * ss - vv;
  else if (vv < ss)
    return 10000000000;
}
int main() {
  int n, k, s, t;
  int time;
  int ans[200005];
  int temp;
  scanf("%d%d%d%d", &n, &k, &s, &t);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &car[i].c, &car[i].v);
  }
  for (int i = 1; i <= k; i++) {
    scanf("%d", &gas[i]);
  }
  gas[0] = 0;
  gas[k + 1] = s;
  sort(gas, gas + k + 2);
  temp = 0;
  for (int i = 0; i < n; i++) {
    int sum = 0;
    for (int j = 1; j <= k + 1; j++) {
      time = Time(j, car[i].v);
      printf("第%d段用的时间为%d\n", j, time);
      sum += time;
    }
    printf("第%d辆车用的时间为%d\n", i, sum);
    if (sum <= t) ans[temp++] = car[i].c;
  }
  sort(ans, ans + temp);
  for (int i = 0; i < temp; i++) {
    printf("%d\n", ans[i]);
  }
  if (ans[0] == 10000000000)
    printf("-1\n");
  else
    printf("%d\n", ans[0]);
  return 0;
}
