#include <bits/stdc++.h>
using namespace std;
int dist[605][605], u, v, a[205], n, Vv, E, m, fix[605], match[605], w, ans, l,
    r, k;
priority_queue<pair<int, int>, vector<pair<int, int> >,
               greater<pair<int, int> > >
    q;
vector<pair<int, int> > V[605];
vector<pair<pair<int, int>, int> > ed;
void dijkstra(int cur) {
  for (int i = 1; i <= Vv; i++) dist[cur][i] = 1e7;
  q.push({0, cur});
  dist[cur][cur] = 0;
  while (q.size()) {
    u = q.top().second;
    int d = q.top().first;
    q.pop();
    if (d > dist[cur][u]) continue;
    for (int i = 0; i < V[u].size(); i++) {
      v = V[u][i].first;
      if (dist[cur][v] > dist[cur][u] + V[u][i].second) {
        dist[cur][v] = dist[cur][u] + V[u][i].second;
        q.push({dist[cur][v], v});
      }
    }
  }
}
bool dfs(int u, int T) {
  for (int v = 1; v <= Vv; v++) {
    if (!fix[v] && dist[u][v] <= T) {
      fix[v] = 1;
      if (match[v] == -1 || dfs(match[v], T)) {
        match[v] = u;
        return true;
      }
    }
  }
  return false;
}
bool check(int T) {
  int pas = 0;
  for (k = 1; k <= Vv; k++) {
    match[k] = -1;
  }
  for (k = 1; k <= n; k++) {
    for (int i = 1; i <= Vv; i++) fix[i] = 0;
    if (dfs(a[k], T)) pas++;
  }
  if (pas >= m) return true;
  return false;
}
int main() {
  cin >> Vv >> E >> n >> m;
  for (k = 1; k <= n; k++) {
    cin >> a[k];
  }
  for (k = 1; k <= E; k++) {
    cin >> u >> v >> w;
    ed.push_back({{u, v}, w});
    V[u].push_back({v, w});
    V[v].push_back({u, w});
  }
  for (k = 1; k <= n; k++) {
    dijkstra(a[k]);
  }
  l = 0;
  r = 1731311;
  ans = -1;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (check(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << ans << endl;
}

### Prompt

Please formulate a Cpp solution to 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;
int dist[605][605], u, v, a[205], n, Vv, E, m, fix[605], match[605], w, ans, l,
    r, k;
priority_queue<pair<int, int>, vector<pair<int, int> >,
               greater<pair<int, int> > >
    q;
vector<pair<int, int> > V[605];
vector<pair<pair<int, int>, int> > ed;
void dijkstra(int cur) {
  for (int i = 1; i <= Vv; i++) dist[cur][i] = 1e7;
  q.push({0, cur});
  dist[cur][cur] = 0;
  while (q.size()) {
    u = q.top().second;
    int d = q.top().first;
    q.pop();
    if (d > dist[cur][u]) continue;
    for (int i = 0; i < V[u].size(); i++) {
      v = V[u][i].first;
      if (dist[cur][v] > dist[cur][u] + V[u][i].second) {
        dist[cur][v] = dist[cur][u] + V[u][i].second;
        q.push({dist[cur][v], v});
      }
    }
  }
}
bool dfs(int u, int T) {
  for (int v = 1; v <= Vv; v++) {
    if (!fix[v] && dist[u][v] <= T) {
      fix[v] = 1;
      if (match[v] == -1 || dfs(match[v], T)) {
        match[v] = u;
        return true;
      }
    }
  }
  return false;
}
bool check(int T) {
  int pas = 0;
  for (k = 1; k <= Vv; k++) {
    match[k] = -1;
  }
  for (k = 1; k <= n; k++) {
    for (int i = 1; i <= Vv; i++) fix[i] = 0;
    if (dfs(a[k], T)) pas++;
  }
  if (pas >= m) return true;
  return false;
}
int main() {
  cin >> Vv >> E >> n >> m;
  for (k = 1; k <= n; k++) {
    cin >> a[k];
  }
  for (k = 1; k <= E; k++) {
    cin >> u >> v >> w;
    ed.push_back({{u, v}, w});
    V[u].push_back({v, w});
    V[v].push_back({u, w});
  }
  for (k = 1; k <= n; k++) {
    dijkstra(a[k]);
  }
  l = 0;
  r = 1731311;
  ans = -1;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (check(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << ans << endl;
}

```