#include <bits/stdc++.h>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const int INF = 2000000000;
const long long LINF = 2000000000000000000;
template <typename T>
void print(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) cout << a[i] << ' ';
  cout << '\n';
}
template <typename T>
void print(deque<T>& a) {
  for (int i = 0; i < a.size(); i++) cout << a[i] << ' ';
  cout << '\n';
}
template <typename T>
void print(vector<vector<T>>& a) {
  for (int i = 0; i < a.size(); i++) {
    for (int j = 0; j < a[i].size(); j++) cout << a[i][j] << ' ';
    cout << '\n';
  }
}
template <typename T>
void input(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) cin >> a[i];
}
template <typename T>
void input(deque<T>& a) {
  for (int i = 0; i < a.size(); i++) cin >> a[i];
}
template <typename T>
void input(vector<vector<T>>& a) {
  for (int i = 0; i < a.size(); i++)
    for (int j = 0; j < a[i].size(); j++) cin >> a[i][j];
}
vector<vector<pair<long long, long long>>> gg;
vector<vector<int>> g;
vector<int> r;
vector<bool> used;
int V, E, N, K;
long long t;
vector<long long> d;
int cur = 0;
void dijkstra(int s) {
  fill((d).begin(), (d).end(), LINF);
  d[s] = 0;
  set<pair<long long, long long>> q;
  q.insert({d[s], s});
  while (!q.empty()) {
    int v = q.begin()->second;
    q.erase(q.begin());
    for (auto to : gg[v])
      if (d[v] + to.second < d[to.first]) {
        q.erase({d[to.first], to.first});
        d[to.first] = d[v] + to.second;
        q.insert({d[to.first], to.first});
      }
  }
  for (int v = 0; v < V; v++)
    if (d[v] <= t) g[cur].push_back(v);
}
bool dfs(int v) {
  if (used[v]) return false;
  used[v] = true;
  for (auto to : g[v])
    if (r[to] == -1 || dfs(r[to])) {
      r[to] = v;
      return true;
    }
  return false;
}
int kuhn() {
  int ans = 0;
  fill((r).begin(), (r).end(), -1);
  fill((used).begin(), (used).end(), false);
  for (int v = 0; v < N; v++) {
    if (dfs(v)) {
      fill((used).begin(), (used).end(), false);
      ans++;
    }
  }
  return ans;
}
void solve() {
  cin >> V >> E >> N >> K;
  gg = vector<vector<pair<long long, long long>>>(V);
  d = vector<long long>(V);
  r = vector<int>(V, -1);
  used = vector<bool>(N);
  vector<int> type(N);
  input(type);
  for (auto& to : type) to--;
  for (int i = 0; i < E; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    gg[u].push_back({v, w});
    gg[v].push_back({u, w});
  }
  long long left = -1, right = 2 * 1731311 + 1;
  vector<int> done(V, -1);
  while (right - left > 1) {
    long long mid = (left + right) / 2;
    g = vector<vector<int>>(N);
    cur = 0;
    fill((done).begin(), (done).end(), -1);
    for (int v = 0; v < N; v++) {
      t = mid;
      if (done[type[v]] != -1)
        for (auto to : g[done[type[v]]]) g[v].push_back(to);
      else {
        dijkstra(type[v]);
        done[type[v]] = v;
      }
      cur++;
    }
    int res = kuhn();
    if (right > 1731311 && res < K) {
      cout << -1;
      return;
    }
    if (res >= K)
      right = mid;
    else
      left = mid;
  }
  cout << right;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tst = 1;
  while (tst--) solve();
}

### Prompt

Please create a solution in CPP 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 warning(disable : 4996)
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const int INF = 2000000000;
const long long LINF = 2000000000000000000;
template <typename T>
void print(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) cout << a[i] << ' ';
  cout << '\n';
}
template <typename T>
void print(deque<T>& a) {
  for (int i = 0; i < a.size(); i++) cout << a[i] << ' ';
  cout << '\n';
}
template <typename T>
void print(vector<vector<T>>& a) {
  for (int i = 0; i < a.size(); i++) {
    for (int j = 0; j < a[i].size(); j++) cout << a[i][j] << ' ';
    cout << '\n';
  }
}
template <typename T>
void input(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) cin >> a[i];
}
template <typename T>
void input(deque<T>& a) {
  for (int i = 0; i < a.size(); i++) cin >> a[i];
}
template <typename T>
void input(vector<vector<T>>& a) {
  for (int i = 0; i < a.size(); i++)
    for (int j = 0; j < a[i].size(); j++) cin >> a[i][j];
}
vector<vector<pair<long long, long long>>> gg;
vector<vector<int>> g;
vector<int> r;
vector<bool> used;
int V, E, N, K;
long long t;
vector<long long> d;
int cur = 0;
void dijkstra(int s) {
  fill((d).begin(), (d).end(), LINF);
  d[s] = 0;
  set<pair<long long, long long>> q;
  q.insert({d[s], s});
  while (!q.empty()) {
    int v = q.begin()->second;
    q.erase(q.begin());
    for (auto to : gg[v])
      if (d[v] + to.second < d[to.first]) {
        q.erase({d[to.first], to.first});
        d[to.first] = d[v] + to.second;
        q.insert({d[to.first], to.first});
      }
  }
  for (int v = 0; v < V; v++)
    if (d[v] <= t) g[cur].push_back(v);
}
bool dfs(int v) {
  if (used[v]) return false;
  used[v] = true;
  for (auto to : g[v])
    if (r[to] == -1 || dfs(r[to])) {
      r[to] = v;
      return true;
    }
  return false;
}
int kuhn() {
  int ans = 0;
  fill((r).begin(), (r).end(), -1);
  fill((used).begin(), (used).end(), false);
  for (int v = 0; v < N; v++) {
    if (dfs(v)) {
      fill((used).begin(), (used).end(), false);
      ans++;
    }
  }
  return ans;
}
void solve() {
  cin >> V >> E >> N >> K;
  gg = vector<vector<pair<long long, long long>>>(V);
  d = vector<long long>(V);
  r = vector<int>(V, -1);
  used = vector<bool>(N);
  vector<int> type(N);
  input(type);
  for (auto& to : type) to--;
  for (int i = 0; i < E; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    gg[u].push_back({v, w});
    gg[v].push_back({u, w});
  }
  long long left = -1, right = 2 * 1731311 + 1;
  vector<int> done(V, -1);
  while (right - left > 1) {
    long long mid = (left + right) / 2;
    g = vector<vector<int>>(N);
    cur = 0;
    fill((done).begin(), (done).end(), -1);
    for (int v = 0; v < N; v++) {
      t = mid;
      if (done[type[v]] != -1)
        for (auto to : g[done[type[v]]]) g[v].push_back(to);
      else {
        dijkstra(type[v]);
        done[type[v]] = v;
      }
      cur++;
    }
    int res = kuhn();
    if (right > 1731311 && res < K) {
      cout << -1;
      return;
    }
    if (res >= K)
      right = mid;
    else
      left = mid;
  }
  cout << right;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tst = 1;
  while (tst--) solve();
}

```