#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const int maxn = 600, maxk = 200;
struct edge {
  int to, w;
  edge(int to, int w) : to(to), w(w) {}
};
int n, m, k, r;
int start[maxk];
vector<edge> graph[maxn];
int dist[maxk][maxn], used[maxn], to[maxn];
vector<int> g[maxk];
int lijwerferlkjre;
bool go(int v) {
  if (used[v] == lijwerferlkjre) return false;
  used[v] = lijwerferlkjre;
  for (int u : g[v]) {
    if (to[u] == -1 || go(to[u])) {
      to[u] = v;
      return true;
    }
  }
  return false;
}
bool check(int T) {
  for (int i = 0; i < k; ++i) {
    g[i].clear();
    for (int j = 0; j < n; ++j) {
      if (dist[i][j] <= T) {
        g[i].emplace_back(j);
      }
    }
  }
  fill(to, to + n, -1);
  fill(used, used + n, 0);
  lijwerferlkjre = 0;
  int cnt = 0;
  for (int i = 0; i < k; ++i) {
    ++lijwerferlkjre;
    if (go(i)) ++cnt;
  }
  return cnt >= r;
}
void dijkstra(int s, int *dist) {
  fill(dist, dist + n, inf);
  fill(used, used + n, 0);
  dist[s] = 0;
  for (int it = 0; it < n; ++it) {
    int v = -1;
    for (int i = 0; i < n; ++i) {
      if (used[i]) continue;
      if (v == -1 || dist[i] < dist[v]) v = i;
    }
    if (v == -1) break;
    used[v] = 1;
    for (const auto &e : graph[v]) {
      dist[e.to] = min(dist[e.to], dist[v] + e.w);
    }
  }
}
int main() {
  srand(131);
  ios_base::sync_with_stdio(false);
  cin >> n >> m >> k >> r;
  for (int i = 0; i < k; ++i) {
    cin >> start[i];
    --start[i];
  }
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    --u, --v;
    graph[u].emplace_back(v, w);
    graph[v].emplace_back(u, w);
  }
  for (int i = 0; i < k; ++i) {
    dijkstra(start[i], dist[i]);
  }
  int left = 0, right = 1731312;
  while (left < right) {
    int m = (left + right) / 2;
    if (check(m))
      right = m;
    else
      left = m + 1;
  }
  cout << (left > 1731311 ? -1 : left) << endl;
  return 0;
}

### Prompt

Generate 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 = 1e9;
const int maxn = 600, maxk = 200;
struct edge {
  int to, w;
  edge(int to, int w) : to(to), w(w) {}
};
int n, m, k, r;
int start[maxk];
vector<edge> graph[maxn];
int dist[maxk][maxn], used[maxn], to[maxn];
vector<int> g[maxk];
int lijwerferlkjre;
bool go(int v) {
  if (used[v] == lijwerferlkjre) return false;
  used[v] = lijwerferlkjre;
  for (int u : g[v]) {
    if (to[u] == -1 || go(to[u])) {
      to[u] = v;
      return true;
    }
  }
  return false;
}
bool check(int T) {
  for (int i = 0; i < k; ++i) {
    g[i].clear();
    for (int j = 0; j < n; ++j) {
      if (dist[i][j] <= T) {
        g[i].emplace_back(j);
      }
    }
  }
  fill(to, to + n, -1);
  fill(used, used + n, 0);
  lijwerferlkjre = 0;
  int cnt = 0;
  for (int i = 0; i < k; ++i) {
    ++lijwerferlkjre;
    if (go(i)) ++cnt;
  }
  return cnt >= r;
}
void dijkstra(int s, int *dist) {
  fill(dist, dist + n, inf);
  fill(used, used + n, 0);
  dist[s] = 0;
  for (int it = 0; it < n; ++it) {
    int v = -1;
    for (int i = 0; i < n; ++i) {
      if (used[i]) continue;
      if (v == -1 || dist[i] < dist[v]) v = i;
    }
    if (v == -1) break;
    used[v] = 1;
    for (const auto &e : graph[v]) {
      dist[e.to] = min(dist[e.to], dist[v] + e.w);
    }
  }
}
int main() {
  srand(131);
  ios_base::sync_with_stdio(false);
  cin >> n >> m >> k >> r;
  for (int i = 0; i < k; ++i) {
    cin >> start[i];
    --start[i];
  }
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    --u, --v;
    graph[u].emplace_back(v, w);
    graph[v].emplace_back(u, w);
  }
  for (int i = 0; i < k; ++i) {
    dijkstra(start[i], dist[i]);
  }
  int left = 0, right = 1731312;
  while (left < right) {
    int m = (left + right) / 2;
    if (check(m))
      right = m;
    else
      left = m + 1;
  }
  cout << (left > 1731311 ? -1 : left) << endl;
  return 0;
}

```