#include <bits/stdc++.h>
using namespace std;
const int N = 1001;
const int INF = (int)1e9;
int cap[N][N], flow[N][N];
int trace[N], visited[N];
vector<int> g[N];
void addEdge(int u, int v, int _cap) {
  g[u].push_back(v), g[v].push_back(u);
  cap[u][v] = _cap;
}
int src, snk, res = 0, maxFlow = 0;
void dfs(int u) {
  visited[u] = 1;
  for (int i = 0; i < (int)g[u].size(); ++i) {
    int v = g[u][i];
    if (!visited[v] && cap[u][v] > flow[u][v]) {
      dfs(v);
      trace[v] = u;
    }
  }
}
bool findPath() {
  for (int i = src; i <= snk; ++i) {
    visited[i] = 0;
  }
  dfs(src);
  return visited[snk];
}
void incFlow() {
  int v = snk, delta = INF;
  while (v != src) {
    int u = trace[v];
    delta = min(delta, cap[u][v] - flow[u][v]);
    v = u;
  }
  v = snk;
  while (v != src) {
    int u = trace[v];
    flow[u][v] += delta, flow[v][u] -= delta;
    v = u;
  }
  maxFlow += delta;
}
int V, E, n, k, cnt[N], dist[601][601];
vector<pair<int, int> > adj[N];
priority_queue<pair<int, int>, vector<pair<int, int> >,
               greater<pair<int, int> > >
    pq;
void dijkstra(int r) {
  for (int i = 1; i <= V; ++i) dist[r][i] = INF;
  dist[r][r] = 0, pq.push(make_pair(0, r));
  while (pq.size()) {
    int u = pq.top().second, c = pq.top().first;
    pq.pop();
    if (dist[r][u] != c) continue;
    for (int i = 0; i < (int)adj[u].size(); ++i) {
      int v = adj[u][i].first;
      int nc = c + adj[u][i].second;
      if (dist[r][v] > nc) {
        dist[r][v] = nc;
        pq.push(make_pair(nc, v));
      }
    }
  }
}
void inp() {
  cin >> V >> E >> n >> k;
  for (int i = 1; i <= n; ++i) {
    int u;
    cin >> u;
    ++cnt[u];
  }
  for (int i = 1; i <= E; ++i) {
    int u, v, c;
    cin >> u >> v >> c;
    adj[u].push_back(make_pair(v, c));
    adj[v].push_back(make_pair(u, c));
  }
}
int m = 0;
bool check(int mid) {
  for (int i = src; i <= snk; ++i) {
    for (int j = src; j <= snk; ++j) {
      flow[i][j] = 0;
    }
  }
  int pos = 0;
  for (int i = 1; i <= V; ++i) {
    if (cnt[i] == 0) continue;
    ++pos;
    for (int j = 1; j <= V; ++j) {
      if (dist[i][j] <= mid)
        cap[pos][j + m] = INF;
      else
        cap[pos][j + m] = 0;
    }
  }
  maxFlow = 0;
  while (findPath()) {
    incFlow();
  }
  return (maxFlow == k);
}
void solve() {
  for (int i = 1; i <= V; ++i) {
    if (cnt[i] >= 1) {
      dijkstra(i);
      ++m;
    }
  }
  if (k <= m) {
    cout << 0;
    return;
  }
  src = 0, snk = m + V + 2;
  int pos = 0;
  for (int i = 1; i <= V; ++i) {
    addEdge(i + m, snk - 1, 1);
    if (cnt[i] == 0) continue;
    ++pos;
    for (int j = 1; j <= V; ++j) {
      if (dist[i][j] == INF) continue;
      addEdge(pos, j + m, INF);
    }
    addEdge(src, pos, cnt[i]);
  }
  addEdge(snk - 1, snk, k);
  int l = 1, r = 1731311, res = -1;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (check(mid)) {
      res = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  inp();
  solve();
  return 0;
}

### Prompt

Generate 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 N = 1001;
const int INF = (int)1e9;
int cap[N][N], flow[N][N];
int trace[N], visited[N];
vector<int> g[N];
void addEdge(int u, int v, int _cap) {
  g[u].push_back(v), g[v].push_back(u);
  cap[u][v] = _cap;
}
int src, snk, res = 0, maxFlow = 0;
void dfs(int u) {
  visited[u] = 1;
  for (int i = 0; i < (int)g[u].size(); ++i) {
    int v = g[u][i];
    if (!visited[v] && cap[u][v] > flow[u][v]) {
      dfs(v);
      trace[v] = u;
    }
  }
}
bool findPath() {
  for (int i = src; i <= snk; ++i) {
    visited[i] = 0;
  }
  dfs(src);
  return visited[snk];
}
void incFlow() {
  int v = snk, delta = INF;
  while (v != src) {
    int u = trace[v];
    delta = min(delta, cap[u][v] - flow[u][v]);
    v = u;
  }
  v = snk;
  while (v != src) {
    int u = trace[v];
    flow[u][v] += delta, flow[v][u] -= delta;
    v = u;
  }
  maxFlow += delta;
}
int V, E, n, k, cnt[N], dist[601][601];
vector<pair<int, int> > adj[N];
priority_queue<pair<int, int>, vector<pair<int, int> >,
               greater<pair<int, int> > >
    pq;
void dijkstra(int r) {
  for (int i = 1; i <= V; ++i) dist[r][i] = INF;
  dist[r][r] = 0, pq.push(make_pair(0, r));
  while (pq.size()) {
    int u = pq.top().second, c = pq.top().first;
    pq.pop();
    if (dist[r][u] != c) continue;
    for (int i = 0; i < (int)adj[u].size(); ++i) {
      int v = adj[u][i].first;
      int nc = c + adj[u][i].second;
      if (dist[r][v] > nc) {
        dist[r][v] = nc;
        pq.push(make_pair(nc, v));
      }
    }
  }
}
void inp() {
  cin >> V >> E >> n >> k;
  for (int i = 1; i <= n; ++i) {
    int u;
    cin >> u;
    ++cnt[u];
  }
  for (int i = 1; i <= E; ++i) {
    int u, v, c;
    cin >> u >> v >> c;
    adj[u].push_back(make_pair(v, c));
    adj[v].push_back(make_pair(u, c));
  }
}
int m = 0;
bool check(int mid) {
  for (int i = src; i <= snk; ++i) {
    for (int j = src; j <= snk; ++j) {
      flow[i][j] = 0;
    }
  }
  int pos = 0;
  for (int i = 1; i <= V; ++i) {
    if (cnt[i] == 0) continue;
    ++pos;
    for (int j = 1; j <= V; ++j) {
      if (dist[i][j] <= mid)
        cap[pos][j + m] = INF;
      else
        cap[pos][j + m] = 0;
    }
  }
  maxFlow = 0;
  while (findPath()) {
    incFlow();
  }
  return (maxFlow == k);
}
void solve() {
  for (int i = 1; i <= V; ++i) {
    if (cnt[i] >= 1) {
      dijkstra(i);
      ++m;
    }
  }
  if (k <= m) {
    cout << 0;
    return;
  }
  src = 0, snk = m + V + 2;
  int pos = 0;
  for (int i = 1; i <= V; ++i) {
    addEdge(i + m, snk - 1, 1);
    if (cnt[i] == 0) continue;
    ++pos;
    for (int j = 1; j <= V; ++j) {
      if (dist[i][j] == INF) continue;
      addEdge(pos, j + m, INF);
    }
    addEdge(src, pos, cnt[i]);
  }
  addEdge(snk - 1, snk, k);
  int l = 1, r = 1731311, res = -1;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (check(mid)) {
      res = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  inp();
  solve();
  return 0;
}

```