#include <bits/stdc++.h>
using namespace std;
struct Edge {
  int u, v;
  long long cap, flow;
  Edge() {}
  Edge(int u, int v, long long cap) : u(u), v(v), cap(cap), flow(0) {}
};
struct Dinic {
  int N;
  vector<Edge> E;
  vector<vector<int> > g;
  vector<int> d, pt;
  Dinic(int N) : N(N), E(0), g(N), d(N), pt(N) {}
  void addEdge(int u, int v, long long cap) {
    if (u != v) {
      E.push_back(Edge(u, v, cap));
      g[u].push_back(E.size() - 1);
      E.push_back(Edge(v, u, 0));
      g[v].push_back(E.size() - 1);
    }
  }
  bool BFS(int S, int T) {
    queue<int> q;
    q.push(S);
    fill(d.begin(), d.end(), N + 1);
    d[S] = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      if (u == T) break;
      for (int k : g[u]) {
        Edge &e = E[k];
        if (e.flow < e.cap && d[e.v] > d[e.u] + 1) {
          d[e.v] = d[e.u] + 1;
          q.push(e.v);
        }
      }
    }
    return d[T] != N + 1;
  }
  long long DFS(int u, int T, long long flow = -1) {
    if (u == T || flow == 0) return flow;
    for (int &i = pt[u]; i < g[u].size(); ++i) {
      Edge &e = E[g[u][i]];
      Edge &oe = E[g[u][i] ^ 1];
      if (d[e.v] == d[e.u] + 1) {
        long long amt = e.cap - e.flow;
        if (flow != -1 && amt > flow) amt = flow;
        if (long long pushed = DFS(e.v, T, amt)) {
          e.flow += pushed;
          oe.flow -= pushed;
          return pushed;
        }
      }
    }
    return 0;
  }
  long long maxFlow(int S, int T) {
    long long total = 0;
    while (BFS(S, T)) {
      fill(pt.begin(), pt.end(), 0);
      while (long long flow = DFS(S, T)) total += flow;
    }
    return total;
  }
};
const int MAXV = 605;
const int INFTY = 1 << 29;
const int MAXANS = 1731312;
int V, E, N, K;
int A[MAXV];
int dist[MAXV][MAXV];
bool check(int t) {
  Dinic dinic(2 + N + V);
  for (int i = 1; i <= N; i++) {
    dinic.addEdge(0, i, 1);
    for (int j = 1; j <= V; j++)
      if (dist[A[i]][j] <= t) dinic.addEdge(i, N + j, 1);
  }
  for (int i = 1; i <= V; i++) dinic.addEdge(N + i, N + V + 1, 1);
  return dinic.maxFlow(0, N + V + 1) >= K;
}
int main() {
  scanf("%d %d %d %d", &V, &E, &N, &K);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  for (int i = 1; i <= V; i++)
    for (int j = i + 1; j <= V; j++) dist[i][j] = dist[j][i] = INFTY;
  for (int i = 0, a, b, c; i < E; i++) {
    scanf("%d %d %d", &a, &b, &c);
    dist[a][b] = min(dist[a][b], c);
    dist[b][a] = min(dist[b][a], c);
  }
  for (int k = 1; k <= V; k++)
    for (int i = 1; i <= V; i++)
      for (int j = 1; j <= V; j++)
        dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
  int lo = 0, hi = MAXANS;
  while (lo < hi) {
    int mid = lo + (hi - lo) / 2;
    if (check(mid))
      hi = mid;
    else
      lo = mid + 1;
  }
  if (lo == MAXANS)
    printf("-1\n");
  else
    printf("%d\n", lo);
  return 0;
}

### Prompt

Your task is to create 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;
struct Edge {
  int u, v;
  long long cap, flow;
  Edge() {}
  Edge(int u, int v, long long cap) : u(u), v(v), cap(cap), flow(0) {}
};
struct Dinic {
  int N;
  vector<Edge> E;
  vector<vector<int> > g;
  vector<int> d, pt;
  Dinic(int N) : N(N), E(0), g(N), d(N), pt(N) {}
  void addEdge(int u, int v, long long cap) {
    if (u != v) {
      E.push_back(Edge(u, v, cap));
      g[u].push_back(E.size() - 1);
      E.push_back(Edge(v, u, 0));
      g[v].push_back(E.size() - 1);
    }
  }
  bool BFS(int S, int T) {
    queue<int> q;
    q.push(S);
    fill(d.begin(), d.end(), N + 1);
    d[S] = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      if (u == T) break;
      for (int k : g[u]) {
        Edge &e = E[k];
        if (e.flow < e.cap && d[e.v] > d[e.u] + 1) {
          d[e.v] = d[e.u] + 1;
          q.push(e.v);
        }
      }
    }
    return d[T] != N + 1;
  }
  long long DFS(int u, int T, long long flow = -1) {
    if (u == T || flow == 0) return flow;
    for (int &i = pt[u]; i < g[u].size(); ++i) {
      Edge &e = E[g[u][i]];
      Edge &oe = E[g[u][i] ^ 1];
      if (d[e.v] == d[e.u] + 1) {
        long long amt = e.cap - e.flow;
        if (flow != -1 && amt > flow) amt = flow;
        if (long long pushed = DFS(e.v, T, amt)) {
          e.flow += pushed;
          oe.flow -= pushed;
          return pushed;
        }
      }
    }
    return 0;
  }
  long long maxFlow(int S, int T) {
    long long total = 0;
    while (BFS(S, T)) {
      fill(pt.begin(), pt.end(), 0);
      while (long long flow = DFS(S, T)) total += flow;
    }
    return total;
  }
};
const int MAXV = 605;
const int INFTY = 1 << 29;
const int MAXANS = 1731312;
int V, E, N, K;
int A[MAXV];
int dist[MAXV][MAXV];
bool check(int t) {
  Dinic dinic(2 + N + V);
  for (int i = 1; i <= N; i++) {
    dinic.addEdge(0, i, 1);
    for (int j = 1; j <= V; j++)
      if (dist[A[i]][j] <= t) dinic.addEdge(i, N + j, 1);
  }
  for (int i = 1; i <= V; i++) dinic.addEdge(N + i, N + V + 1, 1);
  return dinic.maxFlow(0, N + V + 1) >= K;
}
int main() {
  scanf("%d %d %d %d", &V, &E, &N, &K);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  for (int i = 1; i <= V; i++)
    for (int j = i + 1; j <= V; j++) dist[i][j] = dist[j][i] = INFTY;
  for (int i = 0, a, b, c; i < E; i++) {
    scanf("%d %d %d", &a, &b, &c);
    dist[a][b] = min(dist[a][b], c);
    dist[b][a] = min(dist[b][a], c);
  }
  for (int k = 1; k <= V; k++)
    for (int i = 1; i <= V; i++)
      for (int j = 1; j <= V; j++)
        dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
  int lo = 0, hi = MAXANS;
  while (lo < hi) {
    int mid = lo + (hi - lo) / 2;
    if (check(mid))
      hi = mid;
    else
      lo = mid + 1;
  }
  if (lo == MAXANS)
    printf("-1\n");
  else
    printf("%d\n", lo);
  return 0;
}

```