#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
class Hopcroft {
  vector<vector<int>> g;
  vector<int> U, dist;
  int inf = 1e9;
  bool bfs() {
    queue<int> q;
    for (int u : U)
      if (match[u] == nil)
        dist[u] = 0, q.push(u);
      else
        dist[u] = inf;
    dist[nil] = inf;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      if (u != nil)
        for (int v : g[u])
          if (dist[match[v]] == inf) {
            dist[match[v]] = dist[u] + 1;
            q.push(match[v]);
          }
    }
    return (dist[nil] != inf);
  }
  bool dfs(int u) {
    if (u == nil) return true;
    for (int v : g[u])
      if (dist[match[v]] == dist[u] + 1 and dfs(match[v])) {
        match[v] = u, match[u] = v;
        return true;
      }
    dist[u] = inf;
    return false;
  }

 public:
  vector<int> match;
  int nil, isPerfect, matchSize = 0;
  Hopcroft(vector<vector<int>> &gg, vector<int> &UU) {
    g = gg;
    U = UU;
    nil = g.size();
    match.assign(g.size() + 1, nil);
    dist.assign(g.size() + 1, inf);
    while (bfs())
      for (int u : U)
        if (match[u] == nil and dfs(u)) matchSize++;
    isPerfect = (matchSize == U.size() and g.size() == U.size() * 2);
  }
};
int G[610][610];
const int inf = 1e8;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int v, e, n, k;
  cin >> v >> e >> n >> k;
  for (int i = 0; i < v; ++i) {
    for (int j = 0; j < v; ++j) {
      G[i][j] = i == j ? 0 : inf;
    }
  }
  vector<int> C(n);
  for (int l = 0; l < n; ++l) cin >> C[l], C[l]--;
  for (int i = 0; i < e; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    G[u][v] = min(G[u][v], w);
    G[v][u] = min(G[v][u], w);
  }
  for (int k = 0; k < v; ++k) {
    for (int i = 0; i < v; ++i) {
      for (int j = 0; j < v; ++j) {
        G[i][j] = min(G[i][j], G[i][k] + G[k][j]);
      }
    }
  }
  int L = 0, R = 1731311, ans = -1;
  while (L <= R) {
    int M = (L + R) / 2;
    vector<vector<int>> g(n + v);
    vector<int> u(n);
    for (int i = 0; i < n; i++) u[i] = i;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < v; ++j) {
        if (G[C[i]][j] <= M) {
          g[i].push_back(j + n);
          g[j + n].push_back(i);
        }
      }
    }
    Hopcroft hopcroft(g, u);
    int flow = hopcroft.matchSize;
    if (flow >= k) {
      ans = M;
      R = M - 1;
    } else
      L = M + 1;
  }
  cout << ans << "\n";
}

### 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>
#pragma GCC optimize("Ofast")
using namespace std;
class Hopcroft {
  vector<vector<int>> g;
  vector<int> U, dist;
  int inf = 1e9;
  bool bfs() {
    queue<int> q;
    for (int u : U)
      if (match[u] == nil)
        dist[u] = 0, q.push(u);
      else
        dist[u] = inf;
    dist[nil] = inf;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      if (u != nil)
        for (int v : g[u])
          if (dist[match[v]] == inf) {
            dist[match[v]] = dist[u] + 1;
            q.push(match[v]);
          }
    }
    return (dist[nil] != inf);
  }
  bool dfs(int u) {
    if (u == nil) return true;
    for (int v : g[u])
      if (dist[match[v]] == dist[u] + 1 and dfs(match[v])) {
        match[v] = u, match[u] = v;
        return true;
      }
    dist[u] = inf;
    return false;
  }

 public:
  vector<int> match;
  int nil, isPerfect, matchSize = 0;
  Hopcroft(vector<vector<int>> &gg, vector<int> &UU) {
    g = gg;
    U = UU;
    nil = g.size();
    match.assign(g.size() + 1, nil);
    dist.assign(g.size() + 1, inf);
    while (bfs())
      for (int u : U)
        if (match[u] == nil and dfs(u)) matchSize++;
    isPerfect = (matchSize == U.size() and g.size() == U.size() * 2);
  }
};
int G[610][610];
const int inf = 1e8;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int v, e, n, k;
  cin >> v >> e >> n >> k;
  for (int i = 0; i < v; ++i) {
    for (int j = 0; j < v; ++j) {
      G[i][j] = i == j ? 0 : inf;
    }
  }
  vector<int> C(n);
  for (int l = 0; l < n; ++l) cin >> C[l], C[l]--;
  for (int i = 0; i < e; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    G[u][v] = min(G[u][v], w);
    G[v][u] = min(G[v][u], w);
  }
  for (int k = 0; k < v; ++k) {
    for (int i = 0; i < v; ++i) {
      for (int j = 0; j < v; ++j) {
        G[i][j] = min(G[i][j], G[i][k] + G[k][j]);
      }
    }
  }
  int L = 0, R = 1731311, ans = -1;
  while (L <= R) {
    int M = (L + R) / 2;
    vector<vector<int>> g(n + v);
    vector<int> u(n);
    for (int i = 0; i < n; i++) u[i] = i;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < v; ++j) {
        if (G[C[i]][j] <= M) {
          g[i].push_back(j + n);
          g[j + n].push_back(i);
        }
      }
    }
    Hopcroft hopcroft(g, u);
    int flow = hopcroft.matchSize;
    if (flow >= k) {
      ans = M;
      R = M - 1;
    } else
      L = M + 1;
  }
  cout << ans << "\n";
}

```