#include <bits/stdc++.h>
using namespace std;
vector<bool> vis;
vector<int> man;
vector<vector<int>> G;
int dis[605][605];
int N, M;
int v, e, n, k;
int arr[205];
int canMarry(int u) {
  for (auto v : G[u]) {
    if (vis[v]) {
      continue;
    }
    vis[v] = true;
    if (man[v] == -1 || canMarry(man[v])) {
      man[v] = u;
      return 1;
    }
  }
  return 0;
}
int maxMatching() {
  int ans = 0;
  for (int i = 0; i < N; i++) {
    vis.assign(M, false);
    ans += canMarry(i);
  }
  return ans;
}
vector<vector<pair<int, int>>> adj;
bool f(int t) {
  bool isp[v][v];
  for (int i = 0; i < v; i++) {
    for (int j = 0; j < v; j++) {
      isp[i][j] = false;
      if (dis[i][j] <= t) {
        isp[i][j] = true;
      }
    }
  }
  N = n;
  M = v;
  G.clear();
  G.resize(n);
  vis.assign(M, false);
  man.assign(M, -1);
  for (int i = 0; i < n; i++) {
    int node = arr[i];
    for (int j = 0; j < v; j++) {
      if (isp[node][j]) {
        G[i].push_back(j);
      }
    }
  }
  return (maxMatching() >= k);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> v >> e >> n >> k;
  adj.resize(v);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    arr[i]--;
  }
  for (int i = 0; i < e; i++) {
    int u, w, t;
    cin >> u >> w >> t;
    u--;
    w--;
    adj[u].push_back({w, t});
    adj[w].push_back({u, t});
  }
  int lo = 0;
  int hi = 1731311;
  int ans = -1;
  for (int i = 0; i < v; i++) {
    for (int j = 0; j < v; j++) {
      dis[i][j] = 1000000000;
    }
    dis[i][i] = 0;
    set<pair<int, int>> s;
    bool pro[v];
    for (int j = 0; j < v; j++) {
      pro[j] = false;
    }
    s.insert({dis[i][i], i});
    while (!s.empty()) {
      auto it = s.begin();
      int node = (*it).second;
      if (pro[node]) {
        s.erase((*it));
        continue;
      }
      pro[node] = true;
      s.erase((*it));
      for (auto u : adj[node]) {
        if (dis[i][node] + u.second < dis[i][u.first]) {
          dis[i][u.first] = dis[i][node] + u.second;
          s.insert({dis[i][u.first], u.first});
        }
      }
    }
  }
  while (lo <= hi) {
    int mid = (lo + hi) / 2;
    if (f(mid)) {
      ans = mid;
      hi = mid - 1;
    } else {
      lo = mid + 1;
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write 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;
vector<bool> vis;
vector<int> man;
vector<vector<int>> G;
int dis[605][605];
int N, M;
int v, e, n, k;
int arr[205];
int canMarry(int u) {
  for (auto v : G[u]) {
    if (vis[v]) {
      continue;
    }
    vis[v] = true;
    if (man[v] == -1 || canMarry(man[v])) {
      man[v] = u;
      return 1;
    }
  }
  return 0;
}
int maxMatching() {
  int ans = 0;
  for (int i = 0; i < N; i++) {
    vis.assign(M, false);
    ans += canMarry(i);
  }
  return ans;
}
vector<vector<pair<int, int>>> adj;
bool f(int t) {
  bool isp[v][v];
  for (int i = 0; i < v; i++) {
    for (int j = 0; j < v; j++) {
      isp[i][j] = false;
      if (dis[i][j] <= t) {
        isp[i][j] = true;
      }
    }
  }
  N = n;
  M = v;
  G.clear();
  G.resize(n);
  vis.assign(M, false);
  man.assign(M, -1);
  for (int i = 0; i < n; i++) {
    int node = arr[i];
    for (int j = 0; j < v; j++) {
      if (isp[node][j]) {
        G[i].push_back(j);
      }
    }
  }
  return (maxMatching() >= k);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> v >> e >> n >> k;
  adj.resize(v);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    arr[i]--;
  }
  for (int i = 0; i < e; i++) {
    int u, w, t;
    cin >> u >> w >> t;
    u--;
    w--;
    adj[u].push_back({w, t});
    adj[w].push_back({u, t});
  }
  int lo = 0;
  int hi = 1731311;
  int ans = -1;
  for (int i = 0; i < v; i++) {
    for (int j = 0; j < v; j++) {
      dis[i][j] = 1000000000;
    }
    dis[i][i] = 0;
    set<pair<int, int>> s;
    bool pro[v];
    for (int j = 0; j < v; j++) {
      pro[j] = false;
    }
    s.insert({dis[i][i], i});
    while (!s.empty()) {
      auto it = s.begin();
      int node = (*it).second;
      if (pro[node]) {
        s.erase((*it));
        continue;
      }
      pro[node] = true;
      s.erase((*it));
      for (auto u : adj[node]) {
        if (dis[i][node] + u.second < dis[i][u.first]) {
          dis[i][u.first] = dis[i][node] + u.second;
          s.insert({dis[i][u.first], u.first});
        }
      }
    }
  }
  while (lo <= hi) {
    int mid = (lo + hi) / 2;
    if (f(mid)) {
      ans = mid;
      hi = mid - 1;
    } else {
      lo = mid + 1;
    }
  }
  cout << ans;
  return 0;
}

```