#include <bits/stdc++.h>
using namespace std;
const int N = 205, V = 605;
const int inf = 0x3f3f3f3f;
bool vis[V + N];
vector<int> g[N + V];
int v, e, n, k, a[N], d[V][V], lt[N + V], rt[N + V];
void in() {
  memset(d, inf, sizeof d);
  for (int i = 0; i <= v; i++) d[i][i] = 0;
}
void floyd() {
  for (int k = 1; k <= v; k++)
    for (int i = 1; i <= v; i++)
      for (int j = 1; j <= v; j++) d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}
bool match(int node) {
  if (vis[node]) return 0;
  vis[node] = 1;
  for (auto i : g[node]) {
    if (lt[i] == -1 || match(lt[i])) {
      lt[i] = node;
      rt[node] = i;
      return 1;
    }
  }
  return 0;
}
bool chk(int time) {
  for (auto &i : g) i.clear();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= v; j++) {
      if (d[a[i]][j] > time) continue;
      g[i].push_back(j + N);
      g[j + N].push_back(i);
    }
  }
  memset(lt, -1, sizeof lt);
  memset(rt, -1, sizeof rt);
  int matching = 0;
  for (int i = 1; i <= n; i++) {
    memset(vis, 0, sizeof vis);
    matching += match(i);
  }
  return matching >= k;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> v >> e >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  in();
  while (e--) {
    int u, v, c;
    cin >> u >> v >> c;
    d[u][v] = d[v][u] = min(d[u][v], c);
  }
  floyd();
  int s = 0, e = 1.8e6, mid, ans = -1;
  while (s <= e) {
    mid = s + e >> 1;
    if (chk(mid))
      e = mid - 1, ans = mid;
    else
      s = mid + 1;
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in CPP 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 N = 205, V = 605;
const int inf = 0x3f3f3f3f;
bool vis[V + N];
vector<int> g[N + V];
int v, e, n, k, a[N], d[V][V], lt[N + V], rt[N + V];
void in() {
  memset(d, inf, sizeof d);
  for (int i = 0; i <= v; i++) d[i][i] = 0;
}
void floyd() {
  for (int k = 1; k <= v; k++)
    for (int i = 1; i <= v; i++)
      for (int j = 1; j <= v; j++) d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}
bool match(int node) {
  if (vis[node]) return 0;
  vis[node] = 1;
  for (auto i : g[node]) {
    if (lt[i] == -1 || match(lt[i])) {
      lt[i] = node;
      rt[node] = i;
      return 1;
    }
  }
  return 0;
}
bool chk(int time) {
  for (auto &i : g) i.clear();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= v; j++) {
      if (d[a[i]][j] > time) continue;
      g[i].push_back(j + N);
      g[j + N].push_back(i);
    }
  }
  memset(lt, -1, sizeof lt);
  memset(rt, -1, sizeof rt);
  int matching = 0;
  for (int i = 1; i <= n; i++) {
    memset(vis, 0, sizeof vis);
    matching += match(i);
  }
  return matching >= k;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> v >> e >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  in();
  while (e--) {
    int u, v, c;
    cin >> u >> v >> c;
    d[u][v] = d[v][u] = min(d[u][v], c);
  }
  floyd();
  int s = 0, e = 1.8e6, mid, ans = -1;
  while (s <= e) {
    mid = s + e >> 1;
    if (chk(mid))
      e = mid - 1, ans = mid;
    else
      s = mid + 1;
  }
  cout << ans;
  return 0;
}

```