#include <bits/stdc++.h>
using namespace std;
const int inf = 2e8 + 1, maxs = 1731311;
int v, e, n, k;
int st[200], a, b, t;
vector<pair<int, int> > x[600];
int dist[600][600];
set<pair<int, int> > s;
pair<int, int> c;
int lo, hi, mid;
bool bio[200];
int m[600], p;
int f(int y) {
  if (bio[y]) return 0;
  bio[y] = true;
  for (int i = 0; i < v; i++) {
    if (dist[st[y]][i] <= mid && (m[i] == -1 || f(m[i]))) {
      m[i] = y;
      return 1;
    }
  }
  return 0;
}
int main() {
  cin >> v >> e >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> st[i];
    st[i]--;
  }
  for (int i = 0; i < v; i++) x[i].push_back(make_pair(0, i));
  for (int i = 0; i < e; i++) {
    cin >> a >> b >> t;
    a--;
    b--;
    x[a].push_back(make_pair(t, b));
    x[b].push_back(make_pair(t, a));
  }
  for (int i = 0; i < v; i++) {
    for (int j = 0; j < v; j++) {
      if (i == j)
        dist[i][j] = 0;
      else
        dist[i][j] = inf;
    }
    s.insert(make_pair(0, i));
    while (!s.empty()) {
      c = *s.begin();
      s.erase(c);
      for (int j = 1; j < x[c.second].size(); j++) {
        if (dist[i][x[c.second][j].second] >
            dist[i][c.second] + x[c.second][j].first) {
          s.erase(
              make_pair(dist[i][x[c.second][j].second], x[c.second][j].second));
          dist[i][x[c.second][j].second] =
              dist[i][c.second] + x[c.second][j].first;
          s.insert(
              make_pair(dist[i][x[c.second][j].second], x[c.second][j].second));
        }
      }
    }
  }
  lo = 0;
  hi = maxs + 1;
  while (lo < hi) {
    mid = (lo + hi) / 2;
    for (int i = 0; i < v; i++) m[i] = -1;
    p = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) bio[j] = false;
      p += f(i);
    }
    if (p >= k)
      hi = mid;
    else
      lo = mid + 1;
  }
  if (lo <= maxs)
    cout << lo;
  else
    cout << -1;
  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;
const int inf = 2e8 + 1, maxs = 1731311;
int v, e, n, k;
int st[200], a, b, t;
vector<pair<int, int> > x[600];
int dist[600][600];
set<pair<int, int> > s;
pair<int, int> c;
int lo, hi, mid;
bool bio[200];
int m[600], p;
int f(int y) {
  if (bio[y]) return 0;
  bio[y] = true;
  for (int i = 0; i < v; i++) {
    if (dist[st[y]][i] <= mid && (m[i] == -1 || f(m[i]))) {
      m[i] = y;
      return 1;
    }
  }
  return 0;
}
int main() {
  cin >> v >> e >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> st[i];
    st[i]--;
  }
  for (int i = 0; i < v; i++) x[i].push_back(make_pair(0, i));
  for (int i = 0; i < e; i++) {
    cin >> a >> b >> t;
    a--;
    b--;
    x[a].push_back(make_pair(t, b));
    x[b].push_back(make_pair(t, a));
  }
  for (int i = 0; i < v; i++) {
    for (int j = 0; j < v; j++) {
      if (i == j)
        dist[i][j] = 0;
      else
        dist[i][j] = inf;
    }
    s.insert(make_pair(0, i));
    while (!s.empty()) {
      c = *s.begin();
      s.erase(c);
      for (int j = 1; j < x[c.second].size(); j++) {
        if (dist[i][x[c.second][j].second] >
            dist[i][c.second] + x[c.second][j].first) {
          s.erase(
              make_pair(dist[i][x[c.second][j].second], x[c.second][j].second));
          dist[i][x[c.second][j].second] =
              dist[i][c.second] + x[c.second][j].first;
          s.insert(
              make_pair(dist[i][x[c.second][j].second], x[c.second][j].second));
        }
      }
    }
  }
  lo = 0;
  hi = maxs + 1;
  while (lo < hi) {
    mid = (lo + hi) / 2;
    for (int i = 0; i < v; i++) m[i] = -1;
    p = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) bio[j] = false;
      p += f(i);
    }
    if (p >= k)
      hi = mid;
    else
      lo = mid + 1;
  }
  if (lo <= maxs)
    cout << lo;
  else
    cout << -1;
  return 0;
}

```