#include <bits/stdc++.h>
using namespace std;
int dist[610][610];
vector<pair<int, int> > ve[610];
void dij(int x) {
  for (int i = (0); i < (610); i++) dist[x][i] = 1e9;
  dist[x][x] = 0;
  set<pair<int, int> > se;
  se.insert({0, x});
  while (!se.empty()) {
    auto tr = (*se.begin());
    se.erase(tr);
    int a = tr.second;
    for (int i = (0); i < (ve[a].size()); i++) {
      auto ne = ve[a][i];
      int b = ne.first;
      int c = ne.second;
      if (dist[x][b] > dist[x][a] + c) {
        se.erase({dist[x][b], b});
        dist[x][b] = dist[x][a] + c;
        se.insert({dist[x][b], b});
      }
    }
  }
  return;
}
int team[610];
int bio[610];
int par[610];
int v, e, n, k;
int matched[610];
vector<int> edge[610];
vector<int> li;
int match(int x) {
  if (bio[x]) return 0;
  bio[x] = 1;
  for (int i = (0); i < (edge[x].size()); i++) {
    int ne = edge[x][i];
    if (matched[ne] == -1 || match(matched[ne])) {
      matched[ne] = x;
      return 1;
    }
  }
  return 0;
}
bool can(int x) {
  for (int i = (0); i < (600); i++) edge[i].clear();
  for (int i = (0); i < (li.size()); i++) {
    int tr = li[i];
    for (int j = (0); j < (v); j++) {
      if (dist[tr][j] <= x) {
        edge[i].push_back(j);
      }
    }
  }
  int out = 0;
  memset(matched, -1, sizeof matched);
  for (int i = (0); i < (li.size()); i++) {
    memset(bio, 0, sizeof bio);
    out += match(i);
  }
  return (out >= k);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> v >> e >> n >> k;
  for (int i = (0); i < (n); i++) {
    int x;
    cin >> x;
    team[x - 1]++;
    li.push_back(x - 1);
  }
  for (int i = (0); i < (e); i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    ve[a].push_back({b, c});
    ve[b].push_back({a, c});
  }
  for (int i = (0); i < (v); i++) {
    if (team[i]) dij(i);
  }
  int lo = 0, hi = 1731312;
  while (lo < hi) {
    int mi = (lo + hi) / 2;
    if (can(mi)) {
      hi = mi;
    } else {
      lo = mi + 1;
    }
  }
  if (lo == 1731312)
    cout << "-1\n";
  else
    cout << lo << "\n";
  return 0;
}

### 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>
using namespace std;
int dist[610][610];
vector<pair<int, int> > ve[610];
void dij(int x) {
  for (int i = (0); i < (610); i++) dist[x][i] = 1e9;
  dist[x][x] = 0;
  set<pair<int, int> > se;
  se.insert({0, x});
  while (!se.empty()) {
    auto tr = (*se.begin());
    se.erase(tr);
    int a = tr.second;
    for (int i = (0); i < (ve[a].size()); i++) {
      auto ne = ve[a][i];
      int b = ne.first;
      int c = ne.second;
      if (dist[x][b] > dist[x][a] + c) {
        se.erase({dist[x][b], b});
        dist[x][b] = dist[x][a] + c;
        se.insert({dist[x][b], b});
      }
    }
  }
  return;
}
int team[610];
int bio[610];
int par[610];
int v, e, n, k;
int matched[610];
vector<int> edge[610];
vector<int> li;
int match(int x) {
  if (bio[x]) return 0;
  bio[x] = 1;
  for (int i = (0); i < (edge[x].size()); i++) {
    int ne = edge[x][i];
    if (matched[ne] == -1 || match(matched[ne])) {
      matched[ne] = x;
      return 1;
    }
  }
  return 0;
}
bool can(int x) {
  for (int i = (0); i < (600); i++) edge[i].clear();
  for (int i = (0); i < (li.size()); i++) {
    int tr = li[i];
    for (int j = (0); j < (v); j++) {
      if (dist[tr][j] <= x) {
        edge[i].push_back(j);
      }
    }
  }
  int out = 0;
  memset(matched, -1, sizeof matched);
  for (int i = (0); i < (li.size()); i++) {
    memset(bio, 0, sizeof bio);
    out += match(i);
  }
  return (out >= k);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> v >> e >> n >> k;
  for (int i = (0); i < (n); i++) {
    int x;
    cin >> x;
    team[x - 1]++;
    li.push_back(x - 1);
  }
  for (int i = (0); i < (e); i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    ve[a].push_back({b, c});
    ve[b].push_back({a, c});
  }
  for (int i = (0); i < (v); i++) {
    if (team[i]) dij(i);
  }
  int lo = 0, hi = 1731312;
  while (lo < hi) {
    int mi = (lo + hi) / 2;
    if (can(mi)) {
      hi = mi;
    } else {
      lo = mi + 1;
    }
  }
  if (lo == 1731312)
    cout << "-1\n";
  else
    cout << lo << "\n";
  return 0;
}

```