#include <bits/stdc++.h>
using namespace std;
const int N = 600 + 10;
int n, vist[2 * N];
int dis[N][N];
deque<int> path;
vector<vector<pair<int, int>>> v, v2;
int get_path(int start, int end, int val) {
  if (start == end) {
    path.push_front(start);
    return 2e9;
  }
  if (vist[start]) return 0;
  vist[start] = 1;
  for (int i = 0; i <= 2 * n + 1; i++) {
    if (v[start][i].first != 0 && v[start][i].second <= val) {
      int res = get_path(i, end, val);
      if (res != 0) {
        res = min(res, v[start][i].first);
        path.push_front(start);
        return res;
      }
    }
  }
  return 0;
}
int max_flow(int val, int start = 0, int end = 2 * n + 1) {
  int mxflow = 0;
  while (1) {
    memset(vist, 0, sizeof vist);
    path.clear();
    int newflow = get_path(start, end, val);
    if (newflow == 0) return mxflow;
    mxflow += newflow;
    for (int i = 0; i < path.size() - 1; i++) {
      v[path[i]][path[i + 1]].first -= newflow;
      v[path[i + 1]][path[i]].first += newflow;
    }
  }
}
void floyed() {
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int m, t, k;
  cin >> n >> m >> t >> k;
  v2.resize(2 * n + 2, vector<pair<int, int>>(2 * n + 2, {0, 0}));
  for (int i = 0; i < t; i++) {
    int x;
    cin >> x;
    v2[0][x].first++;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) dis[i][j] = ((i == j) ? 0 : 1e9);
  for (int i = 0; i < m; i++) {
    int u, v, c;
    cin >> u >> v >> c;
    dis[u][v] = min(dis[u][v], c);
    dis[v][u] = min(dis[v][u], c);
  }
  floyed();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      v2[i][j + n].second = dis[i][j];
      v2[i][j + n].first = 500;
    }
    v2[i + n][2 * n + 1].first = 1;
  }
  int low = 0, high = 1731311;
  int ans = -1;
  while (low <= high) {
    v = v2;
    int mid = (low + high) / 2;
    if (max_flow(mid) >= k) {
      ans = mid;
      high = mid - 1;
    } else
      low = mid + 1;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
const int N = 600 + 10;
int n, vist[2 * N];
int dis[N][N];
deque<int> path;
vector<vector<pair<int, int>>> v, v2;
int get_path(int start, int end, int val) {
  if (start == end) {
    path.push_front(start);
    return 2e9;
  }
  if (vist[start]) return 0;
  vist[start] = 1;
  for (int i = 0; i <= 2 * n + 1; i++) {
    if (v[start][i].first != 0 && v[start][i].second <= val) {
      int res = get_path(i, end, val);
      if (res != 0) {
        res = min(res, v[start][i].first);
        path.push_front(start);
        return res;
      }
    }
  }
  return 0;
}
int max_flow(int val, int start = 0, int end = 2 * n + 1) {
  int mxflow = 0;
  while (1) {
    memset(vist, 0, sizeof vist);
    path.clear();
    int newflow = get_path(start, end, val);
    if (newflow == 0) return mxflow;
    mxflow += newflow;
    for (int i = 0; i < path.size() - 1; i++) {
      v[path[i]][path[i + 1]].first -= newflow;
      v[path[i + 1]][path[i]].first += newflow;
    }
  }
}
void floyed() {
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int m, t, k;
  cin >> n >> m >> t >> k;
  v2.resize(2 * n + 2, vector<pair<int, int>>(2 * n + 2, {0, 0}));
  for (int i = 0; i < t; i++) {
    int x;
    cin >> x;
    v2[0][x].first++;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) dis[i][j] = ((i == j) ? 0 : 1e9);
  for (int i = 0; i < m; i++) {
    int u, v, c;
    cin >> u >> v >> c;
    dis[u][v] = min(dis[u][v], c);
    dis[v][u] = min(dis[v][u], c);
  }
  floyed();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      v2[i][j + n].second = dis[i][j];
      v2[i][j + n].first = 500;
    }
    v2[i + n][2 * n + 1].first = 1;
  }
  int low = 0, high = 1731311;
  int ans = -1;
  while (low <= high) {
    v = v2;
    int mid = (low + high) / 2;
    if (max_flow(mid) >= k) {
      ans = mid;
      high = mid - 1;
    } else
      low = mid + 1;
  }
  cout << ans << "\n";
  return 0;
}

```