#include <bits/stdc++.h>
using namespace std;
const int N = 8e2 + 5;
long long v, n, e, k, m, a[N], matchTo[N];
long long start[N], Time[N][N];
vector<int> adj[N];
vector<pair<long long, long long>> ad[N];
bool vis[N];
bool dfs(int node) {
  if (vis[node]) return 0;
  vis[node] = 1;
  for (auto ch : adj[node]) {
    if (!matchTo[ch] || dfs(matchTo[ch])) {
      matchTo[ch] = node;
      return 1;
    }
  }
  return 0;
}
void dijs(long long sc) {
  priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
                 greater<pair<long long, long long>>>
      pq;
  long long cost[v + 2];
  for (int i = 0; i < v + 2; i++) cost[i] = 1e15;
  cost[sc] = 0;
  pq.push({0, sc});
  while (pq.size()) {
    long long node = pq.top().second, cst = pq.top().first;
    pq.pop();
    if (cst > cost[node]) continue;
    for (int i = 0; i < ad[node].size(); i++) {
      long long ch = ad[node][i].first, cst2 = ad[node][i].second;
      if (cst + cst2 < cost[ch]) {
        cost[ch] = cst + cst2;
        pq.push({cst + cst2, ch});
      }
    }
  }
  for (int i = 1; i <= v; i++) Time[sc][i] = cost[i];
}
bool check(long long tim) {
  for (int i = 1; i <= n; i++) {
    adj[i].clear();
    for (int j = 1; j <= v; j++) {
      if (Time[start[i]][j] <= tim) {
        adj[i].push_back(n + j);
      }
    }
  }
  memset(matchTo, 0, sizeof(matchTo));
  int match = 0;
  for (int i = 1; i <= n; i++) {
    memset(vis, 0, sizeof(vis));
    match += dfs(i);
  }
  return match >= k;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> v >> e >> n >> k;
  for (int i = 1; i <= n; i++) cin >> start[i];
  for (int i = 1; i <= e; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    ad[x].push_back({y, c}), ad[y].push_back({x, c});
  }
  for (int i = 1; i <= n; i++) {
    dijs(start[i]);
  }
  long long l = 0, h = 1731311, mid, idx = -1;
  while (l <= h) {
    mid = (l + h) / 2;
    if (check(mid)) {
      idx = mid;
      h = mid - 1;
    } else
      l = mid + 1;
  }
  cout << idx << '\n';
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 = 8e2 + 5;
long long v, n, e, k, m, a[N], matchTo[N];
long long start[N], Time[N][N];
vector<int> adj[N];
vector<pair<long long, long long>> ad[N];
bool vis[N];
bool dfs(int node) {
  if (vis[node]) return 0;
  vis[node] = 1;
  for (auto ch : adj[node]) {
    if (!matchTo[ch] || dfs(matchTo[ch])) {
      matchTo[ch] = node;
      return 1;
    }
  }
  return 0;
}
void dijs(long long sc) {
  priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
                 greater<pair<long long, long long>>>
      pq;
  long long cost[v + 2];
  for (int i = 0; i < v + 2; i++) cost[i] = 1e15;
  cost[sc] = 0;
  pq.push({0, sc});
  while (pq.size()) {
    long long node = pq.top().second, cst = pq.top().first;
    pq.pop();
    if (cst > cost[node]) continue;
    for (int i = 0; i < ad[node].size(); i++) {
      long long ch = ad[node][i].first, cst2 = ad[node][i].second;
      if (cst + cst2 < cost[ch]) {
        cost[ch] = cst + cst2;
        pq.push({cst + cst2, ch});
      }
    }
  }
  for (int i = 1; i <= v; i++) Time[sc][i] = cost[i];
}
bool check(long long tim) {
  for (int i = 1; i <= n; i++) {
    adj[i].clear();
    for (int j = 1; j <= v; j++) {
      if (Time[start[i]][j] <= tim) {
        adj[i].push_back(n + j);
      }
    }
  }
  memset(matchTo, 0, sizeof(matchTo));
  int match = 0;
  for (int i = 1; i <= n; i++) {
    memset(vis, 0, sizeof(vis));
    match += dfs(i);
  }
  return match >= k;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> v >> e >> n >> k;
  for (int i = 1; i <= n; i++) cin >> start[i];
  for (int i = 1; i <= e; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    ad[x].push_back({y, c}), ad[y].push_back({x, c});
  }
  for (int i = 1; i <= n; i++) {
    dijs(start[i]);
  }
  long long l = 0, h = 1731311, mid, idx = -1;
  while (l <= h) {
    mid = (l + h) / 2;
    if (check(mid)) {
      idx = mid;
      h = mid - 1;
    } else
      l = mid + 1;
  }
  cout << idx << '\n';
  return 0;
}

```