#include <bits/stdc++.h>
using namespace std;
const int nmax = 6e2 + 42, inf = 1731311;
int v, e, n, k;
int dist[nmax][nmax];
int teams[nmax];
vector<int> adj[nmax];
int matched[nmax * 2];
bool been[nmax * 2];
bool dfs(int node) {
  if (been[node]) return 0;
  been[node] = 1;
  for (auto k : adj[node])
    if (matched[k] == 0 || dfs(matched[k])) {
      matched[k] = node;
      return 1;
    }
  return 0;
}
int kuhn() {
  for (int i = 1; i <= v + n; i++) matched[i] = 0;
  for (int i = 1; i <= n; i++) {
    memset(been, 0, sizeof(been));
    dfs(i);
  }
  int ret = 0;
  for (int i = 1; i <= v; i++)
    if (matched[i + n]) ret++;
  return ret;
}
bool test(int d) {
  for (int i = 1; i <= n; i++) adj[i].clear();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= v; j++)
      if (dist[teams[i]][j] <= d) {
        adj[i].push_back(j + n);
      }
  return kuhn() >= k;
}
int main() {
  scanf("%i%i%i%i", &v, &e, &n, &k);
  for (int i = 1; i <= v; i++)
    for (int j = 1; j <= v; j++) dist[i][j] = inf + 1;
  for (int i = 1; i <= v; i++) dist[i][i] = 0;
  for (int i = 1; i <= n; i++) scanf("%i", &teams[i]);
  int a, b, t;
  for (int i = 1; i <= e; i++) {
    scanf("%i%i%i", &a, &b, &t);
    dist[a][b] = min(dist[a][b], t);
    dist[b][a] = min(dist[b][a], t);
  }
  for (int k = 1; k <= v; k++)
    for (int i = 1; i <= v; i++)
      for (int j = 1; j <= v; j++)
        dist[i][j] = min(dist[i][k] + dist[k][j], dist[i][j]);
  if (test(inf) == 0) {
    printf("-1\n");
    return 0;
  }
  int ok = inf, not_ok = -1;
  while (ok - not_ok > 1) {
    int av = (ok + not_ok) / 2;
    if (test(av))
      ok = av;
    else
      not_ok = av;
  }
  printf("%i\n", ok);
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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 nmax = 6e2 + 42, inf = 1731311;
int v, e, n, k;
int dist[nmax][nmax];
int teams[nmax];
vector<int> adj[nmax];
int matched[nmax * 2];
bool been[nmax * 2];
bool dfs(int node) {
  if (been[node]) return 0;
  been[node] = 1;
  for (auto k : adj[node])
    if (matched[k] == 0 || dfs(matched[k])) {
      matched[k] = node;
      return 1;
    }
  return 0;
}
int kuhn() {
  for (int i = 1; i <= v + n; i++) matched[i] = 0;
  for (int i = 1; i <= n; i++) {
    memset(been, 0, sizeof(been));
    dfs(i);
  }
  int ret = 0;
  for (int i = 1; i <= v; i++)
    if (matched[i + n]) ret++;
  return ret;
}
bool test(int d) {
  for (int i = 1; i <= n; i++) adj[i].clear();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= v; j++)
      if (dist[teams[i]][j] <= d) {
        adj[i].push_back(j + n);
      }
  return kuhn() >= k;
}
int main() {
  scanf("%i%i%i%i", &v, &e, &n, &k);
  for (int i = 1; i <= v; i++)
    for (int j = 1; j <= v; j++) dist[i][j] = inf + 1;
  for (int i = 1; i <= v; i++) dist[i][i] = 0;
  for (int i = 1; i <= n; i++) scanf("%i", &teams[i]);
  int a, b, t;
  for (int i = 1; i <= e; i++) {
    scanf("%i%i%i", &a, &b, &t);
    dist[a][b] = min(dist[a][b], t);
    dist[b][a] = min(dist[b][a], t);
  }
  for (int k = 1; k <= v; k++)
    for (int i = 1; i <= v; i++)
      for (int j = 1; j <= v; j++)
        dist[i][j] = min(dist[i][k] + dist[k][j], dist[i][j]);
  if (test(inf) == 0) {
    printf("-1\n");
    return 0;
  }
  int ok = inf, not_ok = -1;
  while (ok - not_ok > 1) {
    int av = (ok + not_ok) / 2;
    if (test(av))
      ok = av;
    else
      not_ok = av;
  }
  printf("%i\n", ok);
  return 0;
}

```