#include <bits/stdc++.h>
using namespace std;
const int N = 605;
int id[N], e[N][N], E[N][N], used[N], bl[N];
int v, en, n, m, ans = -1;
void Floyd() {
  for (int k = 1; k <= v; k++)
    for (int i = 1; i <= v; i++)
      for (int j = 1; j <= v; j++) e[i][j] = min(e[i][j], e[i][k] + e[k][j]);
}
bool km(int x) {
  for (int i = 1; i <= v; i++) {
    if (E[x][i] && !used[i]) {
      used[i] = 1;
      if (!bl[i] || km(bl[i])) {
        bl[i] = x;
        return 1;
      }
    }
  }
  return 0;
}
bool check(int mid) {
  memset(E, 0, sizeof E);
  memset(bl, 0, sizeof bl);
  int res = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= v; j++) {
      if (e[id[i]][j] <= mid) E[i][j] = 1;
    }
  for (int i = 1; i <= v; i++) {
    memset(used, 0, sizeof used);
    if (km(i)) res++;
  }
  if (res < m) return 0;
  return 1;
}
int main() {
  scanf("%d%d%d%d", &v, &en, &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &id[i]);
  memset(e, 60, sizeof e);
  for (int i = 1; i <= v; i++) e[i][i] = 0;
  for (int i = 1, u, v, w; i <= en; i++) {
    scanf("%d%d%d", &u, &v, &w);
    if (w < e[u][v]) e[u][v] = e[v][u] = w;
  }
  Floyd();
  int l = 0, r = 1e9;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  printf("%d\n", ans);
}

### Prompt

Develop a solution in CPP to the problem described below:

The competitors of Bubble Cup X gathered after the competition and discussed what is the best way to get to know the host country and its cities.

After exploring the map of Serbia for a while, the competitors came up with the following facts: the country has V cities which are indexed with numbers from 1 to V, and there are E bi-directional roads that connect the cites. Each road has a weight (the time needed to cross that road). There are N teams at the Bubble Cup and the competitors came up with the following plan: each of the N teams will start their journey in one of the V cities, and some of the teams share the starting position.

They want to find the shortest time T, such that every team can move in these T minutes, and the number of different cities they end up in is at least K (because they will only get to know the cities they end up in). A team doesn't have to be on the move all the time, if they like it in a particular city, they can stay there and wait for the time to pass.

Please help the competitors to determine the shortest time T so it's possible for them to end up in at least K different cities or print -1 if that is impossible no matter how they move.

Note that there can exist multiple roads between some cities.

Input

The first line contains four integers: V, E, N and K (1 ≤ V ≤ 600, 1 ≤ E ≤ 20000, 1 ≤ N ≤ min(V, 200), 1 ≤ K ≤ N), number of cities, number of roads, number of teams and the smallest number of different cities they need to end up in, respectively.

The second line contains N integers, the cities where the teams start their journey.

Next E lines contain information about the roads in following format: Ai Bi Ti (1 ≤ Ai, Bi ≤ V, 1 ≤ Ti ≤ 10000), which means that there is a road connecting cities Ai and Bi, and you need Ti minutes to cross that road.

Output

Output a single integer that represents the minimal time the teams can move for, such that they end up in at least K different cities or output -1 if there is no solution.

If the solution exists, result will be no greater than 1731311.

Example

Input

6 7 5 4
5 5 2 2 5
1 3 3
1 5 2
1 6 5
2 5 4
2 6 7
3 4 11
3 5 3


Output

3

Note

Three teams start from city 5, and two teams start from city 2. If they agree to move for 3 minutes, one possible situation would be the following: Two teams in city 2, one team in city 5, one team in city 3 , and one team in city 1. And we see that there are four different cities the teams end their journey at.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 605;
int id[N], e[N][N], E[N][N], used[N], bl[N];
int v, en, n, m, ans = -1;
void Floyd() {
  for (int k = 1; k <= v; k++)
    for (int i = 1; i <= v; i++)
      for (int j = 1; j <= v; j++) e[i][j] = min(e[i][j], e[i][k] + e[k][j]);
}
bool km(int x) {
  for (int i = 1; i <= v; i++) {
    if (E[x][i] && !used[i]) {
      used[i] = 1;
      if (!bl[i] || km(bl[i])) {
        bl[i] = x;
        return 1;
      }
    }
  }
  return 0;
}
bool check(int mid) {
  memset(E, 0, sizeof E);
  memset(bl, 0, sizeof bl);
  int res = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= v; j++) {
      if (e[id[i]][j] <= mid) E[i][j] = 1;
    }
  for (int i = 1; i <= v; i++) {
    memset(used, 0, sizeof used);
    if (km(i)) res++;
  }
  if (res < m) return 0;
  return 1;
}
int main() {
  scanf("%d%d%d%d", &v, &en, &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &id[i]);
  memset(e, 60, sizeof e);
  for (int i = 1; i <= v; i++) e[i][i] = 0;
  for (int i = 1, u, v, w; i <= en; i++) {
    scanf("%d%d%d", &u, &v, &w);
    if (w < e[u][v]) e[u][v] = e[v][u] = w;
  }
  Floyd();
  int l = 0, r = 1e9;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  printf("%d\n", ans);
}

```