#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int Mod = 1e9 + 7;
const int MaxN = 1e6 + 5;
vector<pair<int, int> > G[605];
int team[205];
int dis[605][605], vis[605];
int v, e, n, k;
void Heap_Dij(int s, int *dist) {
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      que;
  for (int i = 1; i <= v; i++) dist[i] = INF, vis[i] = 0;
  dist[s] = 0;
  que.push(pair<int, int>(0, s));
  while (!que.empty()) {
    pair<int, int> p = que.top();
    que.pop();
    int u = p.second;
    if (vis[u]) continue;
    vis[u] = true;
    for (int i = 0; i < G[u].size(); i++) {
      pair<int, int> e = G[u][i];
      if (dist[e.first] > dist[u] + e.second) {
        dist[e.first] = dist[u] + e.second;
        que.push(pair<int, int>(dist[e.first], e.first));
      }
    }
  }
}
vector<int> edge[605];
int link[605];
bool DFS(int u) {
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    if (!vis[v]) {
      vis[v] = 1;
      if (link[v] == -1 || DFS(link[v])) {
        link[v] = u;
        return true;
      }
    }
  }
  return false;
}
bool check(int mid) {
  for (int i = 1; i <= v; i++) edge[i].clear(), link[i] = -1, vis[i] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= v; j++) {
      if (dis[team[i]][j] <= mid) edge[i].push_back(j);
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    memset(vis, 0, sizeof(vis));
    if (DFS(i)) ans++;
  }
  if (ans >= k) return true;
  return false;
}
int g[605][605], flag[605];
int32_t main() {
  cin >> v >> e >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> team[i];
    flag[team[i]] = 1;
  }
  memset(g, INF, sizeof(g));
  for (int i = 1; i <= e; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    g[u][v] = min(g[u][v], w);
    g[v][u] = min(g[v][u], w);
  }
  for (int i = 1; i <= v; i++) {
    for (int j = i + 1; j <= v; j++) {
      if (g[i][j] < INF) {
        G[i].push_back(pair<int, int>(j, g[i][j]));
        G[j].push_back(pair<int, int>(i, g[i][j]));
      }
    }
  }
  for (int i = 1; i <= v; i++)
    if (flag[i]) Heap_Dij(i, dis[i]);
  int l = 0, r = 10000000, ans = -1;
  while (l <= r) {
    int mid = l + (r - l) / 2;
    if (check(mid))
      r = mid - 1, ans = mid;
    else
      l = mid + 1;
  }
  cout << ans << endl;
  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;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int Mod = 1e9 + 7;
const int MaxN = 1e6 + 5;
vector<pair<int, int> > G[605];
int team[205];
int dis[605][605], vis[605];
int v, e, n, k;
void Heap_Dij(int s, int *dist) {
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      que;
  for (int i = 1; i <= v; i++) dist[i] = INF, vis[i] = 0;
  dist[s] = 0;
  que.push(pair<int, int>(0, s));
  while (!que.empty()) {
    pair<int, int> p = que.top();
    que.pop();
    int u = p.second;
    if (vis[u]) continue;
    vis[u] = true;
    for (int i = 0; i < G[u].size(); i++) {
      pair<int, int> e = G[u][i];
      if (dist[e.first] > dist[u] + e.second) {
        dist[e.first] = dist[u] + e.second;
        que.push(pair<int, int>(dist[e.first], e.first));
      }
    }
  }
}
vector<int> edge[605];
int link[605];
bool DFS(int u) {
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    if (!vis[v]) {
      vis[v] = 1;
      if (link[v] == -1 || DFS(link[v])) {
        link[v] = u;
        return true;
      }
    }
  }
  return false;
}
bool check(int mid) {
  for (int i = 1; i <= v; i++) edge[i].clear(), link[i] = -1, vis[i] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= v; j++) {
      if (dis[team[i]][j] <= mid) edge[i].push_back(j);
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    memset(vis, 0, sizeof(vis));
    if (DFS(i)) ans++;
  }
  if (ans >= k) return true;
  return false;
}
int g[605][605], flag[605];
int32_t main() {
  cin >> v >> e >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> team[i];
    flag[team[i]] = 1;
  }
  memset(g, INF, sizeof(g));
  for (int i = 1; i <= e; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    g[u][v] = min(g[u][v], w);
    g[v][u] = min(g[v][u], w);
  }
  for (int i = 1; i <= v; i++) {
    for (int j = i + 1; j <= v; j++) {
      if (g[i][j] < INF) {
        G[i].push_back(pair<int, int>(j, g[i][j]));
        G[j].push_back(pair<int, int>(i, g[i][j]));
      }
    }
  }
  for (int i = 1; i <= v; i++)
    if (flag[i]) Heap_Dij(i, dis[i]);
  int l = 0, r = 10000000, ans = -1;
  while (l <= r) {
    int mid = l + (r - l) / 2;
    if (check(mid))
      r = mid - 1, ans = mid;
    else
      l = mid + 1;
  }
  cout << ans << endl;
  return 0;
}

```