#include <bits/stdc++.h>
using namespace std;
const long long inf = (long long)1e18;
const long long mod = (long long)1e9 + 7;
const double pi = acos(-1.0);
const double eps = (double)1e-9;
const int dx[] = {0, 0, 1, 0, -1};
const int dy[] = {0, 1, 0, -1, 0};
const int N = 605;
const int INF = (int)1e9 + 228;
struct Edge {
  int v, to, flow, cap;
};
struct Graph {
  vector<Edge> edges;
  vector<vector<int> > g;
  vector<int> ptr, d;
  queue<int> q;
  int n;
  Graph(){};
  Graph(int _n) : g(_n + 1), ptr(_n + 1), d(_n + 1), n(_n) {}
  void make_edge(int v, int u, int cap) {
    g[v].push_back((int)edges.size());
    edges.push_back({v, u, 0, cap});
    g[u].push_back((int)edges.size());
    edges.push_back({u, v, 0, 0});
  }
  bool bfs() {
    fill(d.begin(), d.end(), -1);
    d[0] = 0;
    q.push(0);
    while (!q.empty()) {
      int v = q.front();
      q.pop();
      for (auto id : g[v]) {
        Edge e = edges[id];
        int to = e.to;
        if (e.flow < e.cap && d[to] == -1) {
          d[to] = d[v] + 1;
          q.push(to);
        }
      }
    }
    return d[n] != -1;
  }
  bool dfs(int v) {
    if (v == n) return true;
    for (int& i = ptr[v]; i < (int)g[v].size(); i++) {
      int id = g[v][i];
      Edge e = edges[id];
      int to = e.to;
      if (d[to] == d[v] + 1 && e.flow < e.cap && dfs(to)) {
        edges[id].flow++, edges[id ^ 1].flow--;
        return true;
      }
    }
    return false;
  }
  int dinic() {
    int res = 0;
    while (bfs()) {
      fill(ptr.begin(), ptr.end(), 0);
      while (dfs(0)) res++;
    }
    return res;
  }
} graph;
int V, E, n, k, cnt, u[N], res[N][N], d[N];
vector<pair<int, int> > g[N];
set<pair<int, int> > st;
vector<int> good;
void dijkstra(int v) {
  for (int i = 1; i <= V; i++) {
    if (i != v)
      d[i] = INF;
    else
      d[i] = 0;
    st.insert({d[i], i});
  }
  while (!st.empty()) {
    pair<int, int> p = *st.begin();
    st.erase(p);
    int v = p.second;
    for (auto e : g[v]) {
      int to = e.first, w = e.second;
      if (d[to] > d[v] + w) {
        st.erase({d[to], to});
        d[to] = d[v] + w;
        st.insert({d[to], to});
      }
    }
  }
}
bool check(int x) {
  graph = Graph(cnt + V + 1);
  for (int i = 0; i < (int)good.size(); i++)
    graph.make_edge(0, i + 1, u[good[i]]);
  for (int i = 1; i <= V; i++) graph.make_edge(i + cnt, V + cnt + 1, 1);
  for (int i = 0; i < (int)good.size(); i++)
    for (int j = 1; j <= V; j++)
      if (res[good[i]][j] <= x) graph.make_edge(i + 1, cnt + j, 1);
  return graph.dinic() >= k;
}
int main() {
  scanf("%d%d%d%d", &V, &E, &n, &k);
  for (int i = 1; i <= n; i++) {
    int v;
    scanf("%d", &v);
    if (!u[v]) cnt++, good.push_back(v);
    u[v]++;
  }
  for (int i = 1; i <= E; i++) {
    int v, u, w;
    scanf("%d%d%d", &v, &u, &w);
    g[v].push_back({u, w});
    g[u].push_back({v, w});
  }
  for (int i = 1; i <= V; i++)
    for (int j = 1; j <= V; j++) res[i][j] = INF;
  for (int i = 1; i <= V; i++) {
    if (u[i] == 0) continue;
    dijkstra(i);
    for (int j = 1; j <= V; j++) res[i][j] = d[j];
  }
  int l = -1, r = 1731312, mid;
  while (l < r - 1) {
    mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  if (r == 1731312)
    cout << -1 << endl;
  else
    cout << r << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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 long long inf = (long long)1e18;
const long long mod = (long long)1e9 + 7;
const double pi = acos(-1.0);
const double eps = (double)1e-9;
const int dx[] = {0, 0, 1, 0, -1};
const int dy[] = {0, 1, 0, -1, 0};
const int N = 605;
const int INF = (int)1e9 + 228;
struct Edge {
  int v, to, flow, cap;
};
struct Graph {
  vector<Edge> edges;
  vector<vector<int> > g;
  vector<int> ptr, d;
  queue<int> q;
  int n;
  Graph(){};
  Graph(int _n) : g(_n + 1), ptr(_n + 1), d(_n + 1), n(_n) {}
  void make_edge(int v, int u, int cap) {
    g[v].push_back((int)edges.size());
    edges.push_back({v, u, 0, cap});
    g[u].push_back((int)edges.size());
    edges.push_back({u, v, 0, 0});
  }
  bool bfs() {
    fill(d.begin(), d.end(), -1);
    d[0] = 0;
    q.push(0);
    while (!q.empty()) {
      int v = q.front();
      q.pop();
      for (auto id : g[v]) {
        Edge e = edges[id];
        int to = e.to;
        if (e.flow < e.cap && d[to] == -1) {
          d[to] = d[v] + 1;
          q.push(to);
        }
      }
    }
    return d[n] != -1;
  }
  bool dfs(int v) {
    if (v == n) return true;
    for (int& i = ptr[v]; i < (int)g[v].size(); i++) {
      int id = g[v][i];
      Edge e = edges[id];
      int to = e.to;
      if (d[to] == d[v] + 1 && e.flow < e.cap && dfs(to)) {
        edges[id].flow++, edges[id ^ 1].flow--;
        return true;
      }
    }
    return false;
  }
  int dinic() {
    int res = 0;
    while (bfs()) {
      fill(ptr.begin(), ptr.end(), 0);
      while (dfs(0)) res++;
    }
    return res;
  }
} graph;
int V, E, n, k, cnt, u[N], res[N][N], d[N];
vector<pair<int, int> > g[N];
set<pair<int, int> > st;
vector<int> good;
void dijkstra(int v) {
  for (int i = 1; i <= V; i++) {
    if (i != v)
      d[i] = INF;
    else
      d[i] = 0;
    st.insert({d[i], i});
  }
  while (!st.empty()) {
    pair<int, int> p = *st.begin();
    st.erase(p);
    int v = p.second;
    for (auto e : g[v]) {
      int to = e.first, w = e.second;
      if (d[to] > d[v] + w) {
        st.erase({d[to], to});
        d[to] = d[v] + w;
        st.insert({d[to], to});
      }
    }
  }
}
bool check(int x) {
  graph = Graph(cnt + V + 1);
  for (int i = 0; i < (int)good.size(); i++)
    graph.make_edge(0, i + 1, u[good[i]]);
  for (int i = 1; i <= V; i++) graph.make_edge(i + cnt, V + cnt + 1, 1);
  for (int i = 0; i < (int)good.size(); i++)
    for (int j = 1; j <= V; j++)
      if (res[good[i]][j] <= x) graph.make_edge(i + 1, cnt + j, 1);
  return graph.dinic() >= k;
}
int main() {
  scanf("%d%d%d%d", &V, &E, &n, &k);
  for (int i = 1; i <= n; i++) {
    int v;
    scanf("%d", &v);
    if (!u[v]) cnt++, good.push_back(v);
    u[v]++;
  }
  for (int i = 1; i <= E; i++) {
    int v, u, w;
    scanf("%d%d%d", &v, &u, &w);
    g[v].push_back({u, w});
    g[u].push_back({v, w});
  }
  for (int i = 1; i <= V; i++)
    for (int j = 1; j <= V; j++) res[i][j] = INF;
  for (int i = 1; i <= V; i++) {
    if (u[i] == 0) continue;
    dijkstra(i);
    for (int j = 1; j <= V; j++) res[i][j] = d[j];
  }
  int l = -1, r = 1731312, mid;
  while (l < r - 1) {
    mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  if (r == 1731312)
    cout << -1 << endl;
  else
    cout << r << endl;
  return 0;
}

```