#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 610;
int d[MAX_N][MAX_N];
const int INF = 1e8;
vector<pair<int, int>> nearest[MAX_N];
bool used[MAX_N];
int curs[MAX_N];
int vers[MAX_N];
int v, e, n, k;
int mt[MAX_N];
bool tryKuhn(int verId) {
  if (used[verId]) {
    return false;
  }
  used[verId] = true;
  int ver = vers[verId];
  for (int i = 0; i <= curs[ver]; i++) {
    int to = nearest[ver][i].second;
    if (mt[to] == -1 || tryKuhn(mt[to])) {
      mt[to] = verId;
      return true;
    }
  }
  return false;
}
bool f(int m, const vector<int> &teams) {
  memset(used, 0, sizeof(used));
  for (int i = 0; i < teams.size(); i++) {
    int t = teams[i];
    for (int j = 0; j < v; j++) {
      if (nearest[t][j].first > m) {
        curs[t] = j - 1;
        vers[i] = t;
        break;
      }
      used[nearest[t][j].second] = true;
      if (j == v - 1) {
        curs[t] = j;
        vers[i] = t;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < v; i++) {
    ans += used[i];
    mt[i] = -1;
  }
  if (ans < k) {
    return false;
  }
  int sum = 0;
  for (int i = 0; i < teams.size(); i++) {
    memset(used, 0, sizeof(used));
    sum += tryKuhn(i);
    if (sum >= k) {
      return true;
    }
  }
  return sum >= k;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> v >> e >> n >> k;
  vector<int> teamsStarts(n);
  for (int i = 0; i < n; i++) {
    cin >> teamsStarts[i];
    teamsStarts[i]--;
  }
  for (int i = 0; i < v; i++) {
    for (int j = 0; j < v; j++) {
      d[i][j] = INF;
    }
    d[i][i] = 0;
  }
  for (int i = 0; i < e; i++) {
    int a, b, x;
    cin >> a >> b >> x;
    a--;
    b--;
    if (a == b) continue;
    if (d[a][b] > x) {
      d[a][b] = x;
      d[b][a] = x;
    }
  }
  for (int k = 1; k < v; k++) {
    for (int i = 0; i < v; i++) {
      int val = (i < k ? d[i][k] : d[k][i]);
      for (int j = i + 1; j < v; j++) {
        d[i][j] = min(d[i][j], val + (k < j ? d[k][j] : d[j][k]));
      }
    }
  }
  for (int i = 0; i < v; i++) {
    for (int j = i; j < v; j++) {
      int dist = d[i][j];
      nearest[j].push_back({dist, i});
      nearest[i].push_back({dist, j});
    }
  }
  for (int i = 0; i < v; i++) {
    sort(nearest[i].begin(), nearest[i].end());
  }
  int l = 0;
  int r = 8e6;
  while (r - l > 1) {
    int m = (r + l) / 2;
    if (f(m, teamsStarts)) {
      r = m;
    } else {
      l = m;
    }
  }
  if (f(l, teamsStarts)) {
    cout << l << endl;
    return 0;
  }
  if (f(r, teamsStarts)) {
    cout << r << endl;
    return 0;
  }
  cout << -1 << endl;
  return 0;
}

### 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>
using namespace std;
const int MAX_N = 610;
int d[MAX_N][MAX_N];
const int INF = 1e8;
vector<pair<int, int>> nearest[MAX_N];
bool used[MAX_N];
int curs[MAX_N];
int vers[MAX_N];
int v, e, n, k;
int mt[MAX_N];
bool tryKuhn(int verId) {
  if (used[verId]) {
    return false;
  }
  used[verId] = true;
  int ver = vers[verId];
  for (int i = 0; i <= curs[ver]; i++) {
    int to = nearest[ver][i].second;
    if (mt[to] == -1 || tryKuhn(mt[to])) {
      mt[to] = verId;
      return true;
    }
  }
  return false;
}
bool f(int m, const vector<int> &teams) {
  memset(used, 0, sizeof(used));
  for (int i = 0; i < teams.size(); i++) {
    int t = teams[i];
    for (int j = 0; j < v; j++) {
      if (nearest[t][j].first > m) {
        curs[t] = j - 1;
        vers[i] = t;
        break;
      }
      used[nearest[t][j].second] = true;
      if (j == v - 1) {
        curs[t] = j;
        vers[i] = t;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < v; i++) {
    ans += used[i];
    mt[i] = -1;
  }
  if (ans < k) {
    return false;
  }
  int sum = 0;
  for (int i = 0; i < teams.size(); i++) {
    memset(used, 0, sizeof(used));
    sum += tryKuhn(i);
    if (sum >= k) {
      return true;
    }
  }
  return sum >= k;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> v >> e >> n >> k;
  vector<int> teamsStarts(n);
  for (int i = 0; i < n; i++) {
    cin >> teamsStarts[i];
    teamsStarts[i]--;
  }
  for (int i = 0; i < v; i++) {
    for (int j = 0; j < v; j++) {
      d[i][j] = INF;
    }
    d[i][i] = 0;
  }
  for (int i = 0; i < e; i++) {
    int a, b, x;
    cin >> a >> b >> x;
    a--;
    b--;
    if (a == b) continue;
    if (d[a][b] > x) {
      d[a][b] = x;
      d[b][a] = x;
    }
  }
  for (int k = 1; k < v; k++) {
    for (int i = 0; i < v; i++) {
      int val = (i < k ? d[i][k] : d[k][i]);
      for (int j = i + 1; j < v; j++) {
        d[i][j] = min(d[i][j], val + (k < j ? d[k][j] : d[j][k]));
      }
    }
  }
  for (int i = 0; i < v; i++) {
    for (int j = i; j < v; j++) {
      int dist = d[i][j];
      nearest[j].push_back({dist, i});
      nearest[i].push_back({dist, j});
    }
  }
  for (int i = 0; i < v; i++) {
    sort(nearest[i].begin(), nearest[i].end());
  }
  int l = 0;
  int r = 8e6;
  while (r - l > 1) {
    int m = (r + l) / 2;
    if (f(m, teamsStarts)) {
      r = m;
    } else {
      l = m;
    }
  }
  if (f(l, teamsStarts)) {
    cout << l << endl;
    return 0;
  }
  if (f(r, teamsStarts)) {
    cout << r << endl;
    return 0;
  }
  cout << -1 << endl;
  return 0;
}

```