#include <bits/stdc++.h>
using namespace std;
struct kuhn {
  int n, m;
  vector<bool> used;
  vector<int> mt, cnt;
  vector<vector<int> > gr;
  kuhn(int n, int m) : n(n), m(m) {
    gr.resize(n);
    used.resize(n);
    mt.resize(m, -1);
    cnt.resize(m);
  }
  inline void add(int v, int u) {
    gr[v].push_back(u);
    ++cnt[u];
  }
  vector<int> ord;
  bool lets_celebrate_and(int v) {
    if (used[v]) return 0;
    used[v] = 1;
    for (auto to : gr[v]) {
      if (mt[to] == -1 || lets_celebrate_and(mt[to])) {
        mt[to] = v;
        return 1;
      }
    }
    return 0;
  }
  int mx() {
    for (int i = 0; i < n; ++i) {
      sort((gr[i]).begin(), (gr[i]).end(),
           [&](const int &a, const int &b) { return (cnt[a] < cnt[b]); });
    }
    ord.resize(n);
    iota((ord).begin(), (ord).end(), 0);
    sort((ord).begin(), (ord).end(), [&](const int &a, const int &b) {
      return (gr[a].size() < gr[b].size());
    });
    for (auto i : ord) {
      lets_celebrate_and(i);
      used.assign(n, 0);
    }
    int ans = 0;
    for (auto i : mt) {
      ans += (i != -1);
    }
    return ans;
  }
};
void solve() {
  int n, m, k, e;
  cin >> n >> m >> k >> e;
  vector<int> a(k);
  for (auto &i : a) cin >> i, --i;
  vector<vector<long long> > d(n, vector<long long>(n, 1e18));
  for (int i = 0; i < m; ++i) {
    int v, u, c;
    cin >> v >> u >> c;
    --v;
    --u;
    d[v][u] = d[u][v] = min(d[v][u], 1LL * c);
  }
  for (int i = 0; i < n; ++i) d[i][i] = 0;
  for (int k = 0; k < n; ++k) {
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
      }
    }
  }
  auto ok = [&](long long x) {
    kuhn c(k, n);
    for (int i = 0; i < k; ++i) {
      for (int j = 0; j < n; ++j) {
        if (d[a[i]][j] > x) continue;
        c.add(i, j);
      }
    }
    return (c.mx() >= e);
  };
  long long l = 0, r = 1734000, mid, ans = -1;
  while (l <= r) {
    mid = l + r >> 1;
    if (ok(mid))
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int q = 1;
  while (q--) solve();
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
struct kuhn {
  int n, m;
  vector<bool> used;
  vector<int> mt, cnt;
  vector<vector<int> > gr;
  kuhn(int n, int m) : n(n), m(m) {
    gr.resize(n);
    used.resize(n);
    mt.resize(m, -1);
    cnt.resize(m);
  }
  inline void add(int v, int u) {
    gr[v].push_back(u);
    ++cnt[u];
  }
  vector<int> ord;
  bool lets_celebrate_and(int v) {
    if (used[v]) return 0;
    used[v] = 1;
    for (auto to : gr[v]) {
      if (mt[to] == -1 || lets_celebrate_and(mt[to])) {
        mt[to] = v;
        return 1;
      }
    }
    return 0;
  }
  int mx() {
    for (int i = 0; i < n; ++i) {
      sort((gr[i]).begin(), (gr[i]).end(),
           [&](const int &a, const int &b) { return (cnt[a] < cnt[b]); });
    }
    ord.resize(n);
    iota((ord).begin(), (ord).end(), 0);
    sort((ord).begin(), (ord).end(), [&](const int &a, const int &b) {
      return (gr[a].size() < gr[b].size());
    });
    for (auto i : ord) {
      lets_celebrate_and(i);
      used.assign(n, 0);
    }
    int ans = 0;
    for (auto i : mt) {
      ans += (i != -1);
    }
    return ans;
  }
};
void solve() {
  int n, m, k, e;
  cin >> n >> m >> k >> e;
  vector<int> a(k);
  for (auto &i : a) cin >> i, --i;
  vector<vector<long long> > d(n, vector<long long>(n, 1e18));
  for (int i = 0; i < m; ++i) {
    int v, u, c;
    cin >> v >> u >> c;
    --v;
    --u;
    d[v][u] = d[u][v] = min(d[v][u], 1LL * c);
  }
  for (int i = 0; i < n; ++i) d[i][i] = 0;
  for (int k = 0; k < n; ++k) {
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
      }
    }
  }
  auto ok = [&](long long x) {
    kuhn c(k, n);
    for (int i = 0; i < k; ++i) {
      for (int j = 0; j < n; ++j) {
        if (d[a[i]][j] > x) continue;
        c.add(i, j);
      }
    }
    return (c.mx() >= e);
  };
  long long l = 0, r = 1734000, mid, ans = -1;
  while (l <= r) {
    mid = l + r >> 1;
    if (ok(mid))
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int q = 1;
  while (q--) solve();
}

```