#include <bits/stdc++.h>
using namespace std;
const int MAX = 1005, INF = (1e9);
int cnt[MAX], v, e, n, k, dis[MAX][MAX], ty[MAX];
namespace hc {
bool visited[MAX];
vector<int> adj[MAX];
int n, L[MAX], R[MAX], Q[MAX], len[MAX], dis[MAX], parent[MAX];
inline void init(int nodes) {
  n = nodes, memset(len, 0, sizeof(len));
  for (int i = 0; i < MAX; i++) adj[i].clear();
}
inline void add_edge(int u, int v) {
  len[u]++;
  adj[u].push_back(v);
}
bool dfs(int i) {
  for (int j = 0; j < len[i]; j++) {
    int x = adj[i][j];
    if (L[x] == -1 || (parent[L[x]] == i)) {
      if (L[x] == -1 || dfs(L[x])) {
        L[x] = i, R[i] = x;
        return true;
      }
    }
  }
  return false;
}
bool bfs() {
  memset(visited, 0, sizeof(visited));
  int i, j, x, d, f = 0, l = 0;
  for (i = 0; i < n; i++) {
    if (R[i] == -1) {
      visited[i] = true;
      Q[l++] = i, dis[i] = 0;
    }
  }
  while (f < l) {
    i = Q[f++];
    for (j = 0; j < len[i]; j++) {
      x = adj[i][j], d = L[x];
      if (d == -1)
        return true;
      else if (!visited[d]) {
        Q[l++] = d;
        parent[d] = i, visited[d] = true, dis[d] = dis[i] + 1;
      }
    }
  }
  return false;
}
int hopcroft_karp() {
  int res = 0;
  memset(L, -1, sizeof(L));
  memset(R, -1, sizeof(R));
  while (bfs()) {
    for (int i = 0; i < n; i++) {
      if (R[i] == -1 && dfs(i)) res++;
    }
  }
  return res;
}
}  // namespace hc
int main() {
  ios::sync_with_stdio(false);
  int u, v, c;
  cin >> v >> e >> n >> k;
  for (int i = 1; i <= v; i++)
    for (int j = 1; j <= v; j++) dis[i][j] = (i == j) ? 0 : INF;
  for (int i = 1; i <= n; i++) cin >> ty[i];
  for (int i = 0; i <= e - 1; i++) {
    int u, v, c;
    cin >> u >> v >> c;
    dis[v][u] = dis[u][v] = min(dis[u][v], c);
  }
  for (int k = 1; k <= v; k++)
    for (int i = 1; i <= v; i++)
      for (int j = 1; j <= v; j++)
        dis[i][j] = min(dis[i][k] + dis[k][j], dis[i][j]);
  int st = 0, en = 1731311, mid, prin = -1;
  while (st <= en) {
    hc::init(n + v);
    mid = (st + en) >> 1;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= v; j++)
        if (dis[ty[i]][j] <= mid) hc::add_edge(i, j);
    if (hc::hopcroft_karp() >= k) {
      en = mid - 1;
      if (prin == -1)
        prin = mid;
      else
        prin = min(prin, mid);
    } else
      st = mid + 1;
  }
  cout << prin << endl;
  return 0;
}

### Prompt

Your challenge is to write 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 MAX = 1005, INF = (1e9);
int cnt[MAX], v, e, n, k, dis[MAX][MAX], ty[MAX];
namespace hc {
bool visited[MAX];
vector<int> adj[MAX];
int n, L[MAX], R[MAX], Q[MAX], len[MAX], dis[MAX], parent[MAX];
inline void init(int nodes) {
  n = nodes, memset(len, 0, sizeof(len));
  for (int i = 0; i < MAX; i++) adj[i].clear();
}
inline void add_edge(int u, int v) {
  len[u]++;
  adj[u].push_back(v);
}
bool dfs(int i) {
  for (int j = 0; j < len[i]; j++) {
    int x = adj[i][j];
    if (L[x] == -1 || (parent[L[x]] == i)) {
      if (L[x] == -1 || dfs(L[x])) {
        L[x] = i, R[i] = x;
        return true;
      }
    }
  }
  return false;
}
bool bfs() {
  memset(visited, 0, sizeof(visited));
  int i, j, x, d, f = 0, l = 0;
  for (i = 0; i < n; i++) {
    if (R[i] == -1) {
      visited[i] = true;
      Q[l++] = i, dis[i] = 0;
    }
  }
  while (f < l) {
    i = Q[f++];
    for (j = 0; j < len[i]; j++) {
      x = adj[i][j], d = L[x];
      if (d == -1)
        return true;
      else if (!visited[d]) {
        Q[l++] = d;
        parent[d] = i, visited[d] = true, dis[d] = dis[i] + 1;
      }
    }
  }
  return false;
}
int hopcroft_karp() {
  int res = 0;
  memset(L, -1, sizeof(L));
  memset(R, -1, sizeof(R));
  while (bfs()) {
    for (int i = 0; i < n; i++) {
      if (R[i] == -1 && dfs(i)) res++;
    }
  }
  return res;
}
}  // namespace hc
int main() {
  ios::sync_with_stdio(false);
  int u, v, c;
  cin >> v >> e >> n >> k;
  for (int i = 1; i <= v; i++)
    for (int j = 1; j <= v; j++) dis[i][j] = (i == j) ? 0 : INF;
  for (int i = 1; i <= n; i++) cin >> ty[i];
  for (int i = 0; i <= e - 1; i++) {
    int u, v, c;
    cin >> u >> v >> c;
    dis[v][u] = dis[u][v] = min(dis[u][v], c);
  }
  for (int k = 1; k <= v; k++)
    for (int i = 1; i <= v; i++)
      for (int j = 1; j <= v; j++)
        dis[i][j] = min(dis[i][k] + dis[k][j], dis[i][j]);
  int st = 0, en = 1731311, mid, prin = -1;
  while (st <= en) {
    hc::init(n + v);
    mid = (st + en) >> 1;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= v; j++)
        if (dis[ty[i]][j] <= mid) hc::add_edge(i, j);
    if (hc::hopcroft_karp() >= k) {
      en = mid - 1;
      if (prin == -1)
        prin = mid;
      else
        prin = min(prin, mid);
    } else
      st = mid + 1;
  }
  cout << prin << endl;
  return 0;
}

```