#include <bits/stdc++.h>
using namespace std;
const int N = 605, M = 205;
int n, m, t, k;
int dist[N][N], city[N];
void floyd() {
  for (int i = 0; i < n; i++) dist[i][i] = 0;
  for (int k = 0; k < n; k++)
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        if (dist[i][k] + dist[k][j] < dist[i][j])
          dist[i][j] = dist[i][k] + dist[k][j];
}
vector<int> adj[M];
int husband[N], vis[M], vid;
bool find_match(int node) {
  if (vis[node] == vid) return 0;
  vis[node] = vid;
  for (int child : adj[node]) {
    if (husband[child] == -1 || find_match(husband[child])) {
      husband[child] = node;
      return 1;
    }
  }
  return 0;
}
int max_match() {
  memset(husband, -1, sizeof husband);
  int cnt = 0;
  for (int i = 0; i < t; i++) {
    ++vid;
    cnt += find_match(i);
  }
  return cnt;
}
void init() {
  for (int i = 0; i < t; i++) adj[i].clear();
}
bool valid(int time) {
  init();
  for (int i = 0; i < t; i++)
    for (int j = 0; j < n; j++)
      if (dist[city[i]][j] <= time) adj[i].push_back(j);
  return max_match() >= k;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m >> t >> k;
  for (int i = 0; i < t; i++) {
    cin >> city[i];
    city[i]--;
  }
  memset(dist, '?', sizeof dist);
  while (m--) {
    int u, v, c;
    cin >> u >> v >> c;
    u--, v--;
    dist[u][v] = min(dist[u][v], c);
    dist[v][u] = min(dist[v][u], c);
  }
  floyd();
  int l = 0, r = 1731311, ans = -1;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (valid(mid)) {
      ans = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  cout << ans;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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, M = 205;
int n, m, t, k;
int dist[N][N], city[N];
void floyd() {
  for (int i = 0; i < n; i++) dist[i][i] = 0;
  for (int k = 0; k < n; k++)
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        if (dist[i][k] + dist[k][j] < dist[i][j])
          dist[i][j] = dist[i][k] + dist[k][j];
}
vector<int> adj[M];
int husband[N], vis[M], vid;
bool find_match(int node) {
  if (vis[node] == vid) return 0;
  vis[node] = vid;
  for (int child : adj[node]) {
    if (husband[child] == -1 || find_match(husband[child])) {
      husband[child] = node;
      return 1;
    }
  }
  return 0;
}
int max_match() {
  memset(husband, -1, sizeof husband);
  int cnt = 0;
  for (int i = 0; i < t; i++) {
    ++vid;
    cnt += find_match(i);
  }
  return cnt;
}
void init() {
  for (int i = 0; i < t; i++) adj[i].clear();
}
bool valid(int time) {
  init();
  for (int i = 0; i < t; i++)
    for (int j = 0; j < n; j++)
      if (dist[city[i]][j] <= time) adj[i].push_back(j);
  return max_match() >= k;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m >> t >> k;
  for (int i = 0; i < t; i++) {
    cin >> city[i];
    city[i]--;
  }
  memset(dist, '?', sizeof dist);
  while (m--) {
    int u, v, c;
    cin >> u >> v >> c;
    u--, v--;
    dist[u][v] = min(dist[u][v], c);
    dist[v][u] = min(dist[v][u], c);
  }
  floyd();
  int l = 0, r = 1731311, ans = -1;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (valid(mid)) {
      ans = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  cout << ans;
}

```