#include <bits/stdc++.h>
using namespace std;
const int maxn = 700;
int V, E, N, K, mark[maxn], mark1[maxn], city[maxn], C[maxn][maxn], match[maxn];
vector<int> G[maxn];
bool khun(int nodo) {
  if (mark1[nodo]) return false;
  mark1[nodo] = 1;
  for (auto i : G[nodo]) {
    if (match[i] == -1 || khun(match[i])) {
      match[i] = nodo;
      return true;
    }
  }
  return false;
}
bool can(int t) {
  for (int i = 1; i <= V; i++) G[i].clear(), match[i] = -1;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= V; j++) {
      if (C[city[i]][j] <= t) G[i].push_back(j);
    }
  }
  int sol = 0;
  for (int i = 1; i <= N; i++) {
    for (int i = 1; i <= V; i++) mark1[i] = 0;
    sol += khun(i);
  }
  return sol >= K;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> V >> E >> N >> K;
  for (int i = 1; i <= N; i++) cin >> city[i];
  for (int i = 1; i <= V; i++)
    for (int j = 1; j <= V; j++) C[i][j] = 1 << 30;
  for (int i = 1; i <= V; i++) C[i][i] = 0;
  int a, b, c;
  for (int i = 1; i <= E; i++) {
    cin >> a >> b >> c;
    C[a][b] = min(C[a][b], c);
    C[b][a] = min(C[b][a], c);
  }
  for (int k = 1; k <= V; k++)
    for (int i = 1; i <= V; i++)
      for (int j = 1; j <= V; j++)
        if (C[i][k] != 1 << 30 && C[k][j] != 1 << 30)
          C[i][j] = min(C[i][j], C[i][k] + C[k][j]);
  int ini = 0, fin = 1731311, sol = 1 << 30;
  while (ini <= fin) {
    int mit = (ini + fin) / 2;
    if (can(mit))
      sol = mit, fin = mit - 1;
    else
      ini = mit + 1;
  }
  if (sol < 1 << 30)
    cout << sol;
  else
    cout << "-1";
  return 0;
}

### Prompt

Your task is to create 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 maxn = 700;
int V, E, N, K, mark[maxn], mark1[maxn], city[maxn], C[maxn][maxn], match[maxn];
vector<int> G[maxn];
bool khun(int nodo) {
  if (mark1[nodo]) return false;
  mark1[nodo] = 1;
  for (auto i : G[nodo]) {
    if (match[i] == -1 || khun(match[i])) {
      match[i] = nodo;
      return true;
    }
  }
  return false;
}
bool can(int t) {
  for (int i = 1; i <= V; i++) G[i].clear(), match[i] = -1;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= V; j++) {
      if (C[city[i]][j] <= t) G[i].push_back(j);
    }
  }
  int sol = 0;
  for (int i = 1; i <= N; i++) {
    for (int i = 1; i <= V; i++) mark1[i] = 0;
    sol += khun(i);
  }
  return sol >= K;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> V >> E >> N >> K;
  for (int i = 1; i <= N; i++) cin >> city[i];
  for (int i = 1; i <= V; i++)
    for (int j = 1; j <= V; j++) C[i][j] = 1 << 30;
  for (int i = 1; i <= V; i++) C[i][i] = 0;
  int a, b, c;
  for (int i = 1; i <= E; i++) {
    cin >> a >> b >> c;
    C[a][b] = min(C[a][b], c);
    C[b][a] = min(C[b][a], c);
  }
  for (int k = 1; k <= V; k++)
    for (int i = 1; i <= V; i++)
      for (int j = 1; j <= V; j++)
        if (C[i][k] != 1 << 30 && C[k][j] != 1 << 30)
          C[i][j] = min(C[i][j], C[i][k] + C[k][j]);
  int ini = 0, fin = 1731311, sol = 1 << 30;
  while (ini <= fin) {
    int mit = (ini + fin) / 2;
    if (can(mit))
      sol = mit, fin = mit - 1;
    else
      ini = mit + 1;
  }
  if (sol < 1 << 30)
    cout << sol;
  else
    cout << "-1";
  return 0;
}

```