#include <bits/stdc++.h>
using namespace std;
int dist[609][609];
vector<pair<int, int>> neigh[609];
void dijkstra(int s) {
  priority_queue<pair<int, int>> Q;
  Q.push({0, s});
  while (!Q.empty()) {
    pair<int, int> cur = Q.top();
    Q.pop();
    cur.first *= -1;
    if (cur.first > dist[s][cur.second]) continue;
    for (auto u : neigh[cur.second]) {
      if (dist[s][u.first] > dist[s][cur.second] + u.second) {
        dist[s][u.first] = dist[s][cur.second] + u.second;
        Q.push({-dist[s][u.first], u.first});
      }
    }
  }
}
vector<int> to[209];
int from[609];
bool vis[209];
bool taken[209];
bool try_kuhn(int v) {
  if (vis[v]) return false;
  vis[v] = true;
  for (int u : to[v]) {
    if (from[u] == -1 || try_kuhn(from[u])) {
      from[u] = v;
      return true;
    }
  }
  return false;
}
int main() {
  int n, e, m, K;
  scanf("%d %d %d %d", &n, &e, &m, &K);
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      dist[j][i] = dist[i][j] = 100000000;
    }
  }
  vector<int> V;
  for (int i = 0; i < m; i++) {
    int s;
    scanf("%d", &s);
    V.push_back(s);
  }
  while (e--) {
    int u, v, x;
    scanf("%d %d %d", &u, &v, &x);
    neigh[u].push_back({v, x});
    neigh[v].push_back({u, x});
  }
  for (int i = 1; i <= n; i++) {
    dijkstra(i);
  }
  int L = 0, R = 1731311;
  int ans = -1;
  while (L <= R) {
    int M = (L + R) / 2;
    for (int i = 1; i <= n; i++) {
      from[i] = -1;
    }
    int cnt = 0;
    for (int i = 1; i <= m; i++) {
      vis[i] = taken[i] = false;
      to[i].clear();
      int u = V[i - 1];
      for (int v = 1; v <= n; v++) {
        if (dist[u][v] <= M) {
          to[i].push_back(v);
          if (!taken[i] && from[v] == -1) {
            taken[i] = true;
            from[v] = i;
            cnt++;
          }
        }
      }
    }
    for (int i = 1; i <= m; i++) {
      if (taken[i]) continue;
      memset(vis, 0, sizeof vis);
      if (try_kuhn(i)) cnt++;
    }
    if (cnt >= K) {
      ans = M;
      R = M - 1;
    } else {
      L = M + 1;
    }
  }
  cout << ans << endl;
}

### Prompt

Please provide a cpp coded solution 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;
int dist[609][609];
vector<pair<int, int>> neigh[609];
void dijkstra(int s) {
  priority_queue<pair<int, int>> Q;
  Q.push({0, s});
  while (!Q.empty()) {
    pair<int, int> cur = Q.top();
    Q.pop();
    cur.first *= -1;
    if (cur.first > dist[s][cur.second]) continue;
    for (auto u : neigh[cur.second]) {
      if (dist[s][u.first] > dist[s][cur.second] + u.second) {
        dist[s][u.first] = dist[s][cur.second] + u.second;
        Q.push({-dist[s][u.first], u.first});
      }
    }
  }
}
vector<int> to[209];
int from[609];
bool vis[209];
bool taken[209];
bool try_kuhn(int v) {
  if (vis[v]) return false;
  vis[v] = true;
  for (int u : to[v]) {
    if (from[u] == -1 || try_kuhn(from[u])) {
      from[u] = v;
      return true;
    }
  }
  return false;
}
int main() {
  int n, e, m, K;
  scanf("%d %d %d %d", &n, &e, &m, &K);
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      dist[j][i] = dist[i][j] = 100000000;
    }
  }
  vector<int> V;
  for (int i = 0; i < m; i++) {
    int s;
    scanf("%d", &s);
    V.push_back(s);
  }
  while (e--) {
    int u, v, x;
    scanf("%d %d %d", &u, &v, &x);
    neigh[u].push_back({v, x});
    neigh[v].push_back({u, x});
  }
  for (int i = 1; i <= n; i++) {
    dijkstra(i);
  }
  int L = 0, R = 1731311;
  int ans = -1;
  while (L <= R) {
    int M = (L + R) / 2;
    for (int i = 1; i <= n; i++) {
      from[i] = -1;
    }
    int cnt = 0;
    for (int i = 1; i <= m; i++) {
      vis[i] = taken[i] = false;
      to[i].clear();
      int u = V[i - 1];
      for (int v = 1; v <= n; v++) {
        if (dist[u][v] <= M) {
          to[i].push_back(v);
          if (!taken[i] && from[v] == -1) {
            taken[i] = true;
            from[v] = i;
            cnt++;
          }
        }
      }
    }
    for (int i = 1; i <= m; i++) {
      if (taken[i]) continue;
      memset(vis, 0, sizeof vis);
      if (try_kuhn(i)) cnt++;
    }
    if (cnt >= K) {
      ans = M;
      R = M - 1;
    } else {
      L = M + 1;
    }
  }
  cout << ans << endl;
}

```