#include <bits/stdc++.h>
using namespace std;
const int N = 666;
bool used[N], occured[N];
int dist[N][N];
int a[N][N];
int V, E, n, k;
int c[N];
inline void dijkstra(int st) {
  if (occured[st]) {
    return;
  }
  occured[st] = 1;
  for (int i = 1; i <= V; i++) {
    used[i] = 0;
    dist[st][i] = 1000000000;
  }
  dist[st][st] = 0;
  int v;
  for (int it = 1; it <= V; it++) {
    v = -1;
    for (int i = 1; i <= V; i++) {
      if (!used[i] && (v == -1 || dist[st][i] < dist[st][v])) {
        v = i;
      }
    }
    used[v] = 1;
    for (int i = 1; i <= V; i++) {
      if (a[v][i] && dist[st][i] > dist[st][v] + a[v][i]) {
        dist[st][i] = dist[st][v] + a[v][i];
      }
    }
  }
}
vector<int> cities;
int mt[N];
bool visited[N];
vector<int> g[N];
inline bool kuhn(int v) {
  if (visited[v]) {
    return false;
  }
  visited[v] = 1;
  for (int i = 0; i < (int)g[v].size(); i++) {
    int to = g[v][i];
    if (mt[to] == -1 || kuhn(mt[to])) {
      mt[to] = v;
      return 1;
    }
  }
  return 0;
}
inline bool check(int val) {
  for (int i = 1; i <= n; i++) {
    g[i].clear();
    for (int j = 1; j <= V; j++) {
      if (dist[c[i]][j] <= val) {
        g[i].push_back(j);
      }
    }
  }
  memset(mt, -1, sizeof(mt));
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    memset(visited, 0, sizeof(visited));
    if (kuhn(i)) {
      ans++;
    }
  }
  return (ans >= k);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> V >> E >> n >> k;
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    if (!used[c[i]]) {
      cities.push_back(c[i]);
      cnt++;
      used[c[i]] = 1;
    }
  }
  int x, y, t;
  for (int i = 1; i <= E; i++) {
    cin >> x >> y >> t;
    if (!a[x][y]) {
      a[x][y] = a[y][x] = 1000000000;
    }
    a[x][y] = min(a[x][y], t);
    a[y][x] = min(a[y][x], t);
  }
  if (cnt >= k) {
    cout << 0;
    return 0;
  }
  for (int v : cities) {
    dijkstra(v);
  }
  int l = 0, r = 1731312, mid;
  while (r - l > 1) {
    mid = (l + r) >> 1;
    if (check(mid)) {
      r = mid;
    } else {
      l = mid;
    }
  }
  if (r == 1731312) {
    cout << "-1";
    return 0;
  }
  cout << r;
  return 0;
}

### Prompt

Create a solution in Cpp for 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 = 666;
bool used[N], occured[N];
int dist[N][N];
int a[N][N];
int V, E, n, k;
int c[N];
inline void dijkstra(int st) {
  if (occured[st]) {
    return;
  }
  occured[st] = 1;
  for (int i = 1; i <= V; i++) {
    used[i] = 0;
    dist[st][i] = 1000000000;
  }
  dist[st][st] = 0;
  int v;
  for (int it = 1; it <= V; it++) {
    v = -1;
    for (int i = 1; i <= V; i++) {
      if (!used[i] && (v == -1 || dist[st][i] < dist[st][v])) {
        v = i;
      }
    }
    used[v] = 1;
    for (int i = 1; i <= V; i++) {
      if (a[v][i] && dist[st][i] > dist[st][v] + a[v][i]) {
        dist[st][i] = dist[st][v] + a[v][i];
      }
    }
  }
}
vector<int> cities;
int mt[N];
bool visited[N];
vector<int> g[N];
inline bool kuhn(int v) {
  if (visited[v]) {
    return false;
  }
  visited[v] = 1;
  for (int i = 0; i < (int)g[v].size(); i++) {
    int to = g[v][i];
    if (mt[to] == -1 || kuhn(mt[to])) {
      mt[to] = v;
      return 1;
    }
  }
  return 0;
}
inline bool check(int val) {
  for (int i = 1; i <= n; i++) {
    g[i].clear();
    for (int j = 1; j <= V; j++) {
      if (dist[c[i]][j] <= val) {
        g[i].push_back(j);
      }
    }
  }
  memset(mt, -1, sizeof(mt));
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    memset(visited, 0, sizeof(visited));
    if (kuhn(i)) {
      ans++;
    }
  }
  return (ans >= k);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> V >> E >> n >> k;
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    if (!used[c[i]]) {
      cities.push_back(c[i]);
      cnt++;
      used[c[i]] = 1;
    }
  }
  int x, y, t;
  for (int i = 1; i <= E; i++) {
    cin >> x >> y >> t;
    if (!a[x][y]) {
      a[x][y] = a[y][x] = 1000000000;
    }
    a[x][y] = min(a[x][y], t);
    a[y][x] = min(a[y][x], t);
  }
  if (cnt >= k) {
    cout << 0;
    return 0;
  }
  for (int v : cities) {
    dijkstra(v);
  }
  int l = 0, r = 1731312, mid;
  while (r - l > 1) {
    mid = (l + r) >> 1;
    if (check(mid)) {
      r = mid;
    } else {
      l = mid;
    }
  }
  if (r == 1731312) {
    cout << "-1";
    return 0;
  }
  cout << r;
  return 0;
}

```