#include <bits/stdc++.h>
using namespace std;
const int N = 700;
const int S = 200;
const int M = N * S + 100;
const int INF = 1000 * 1000 * 1000;
int n1, n2, edges, last[N], previous[M], head[M];
int matching[N], dist[N], Q[N];
bool used[N], vis[N];
inline void init(int _n1, int _n2) {
  n1 = _n1;
  n2 = _n2;
  edges = 0;
  fill(last, last + n1, -1);
}
inline void addEdge(int u, int v) {
  head[edges] = v;
  previous[edges] = last[u];
  last[u] = edges++;
}
inline void bfs() {
  fill(dist, dist + n1, -1);
  int sizeQ = 0;
  for (int u = 0; u < n1; ++u) {
    if (!used[u]) {
      Q[sizeQ++] = u;
      dist[u] = 0;
    }
  }
  for (int i = 0; i < sizeQ; i++) {
    int u1 = Q[i];
    for (int e = last[u1]; e >= 0; e = previous[e]) {
      int u2 = matching[head[e]];
      if (u2 >= 0 && dist[u2] < 0) {
        dist[u2] = dist[u1] + 1;
        Q[sizeQ++] = u2;
      }
    }
  }
}
inline bool dfs(int u1) {
  vis[u1] = true;
  for (int e = last[u1]; e >= 0; e = previous[e]) {
    int v = head[e];
    int u2 = matching[v];
    if (u2 < 0 || !vis[u2] && dist[u2] == dist[u1] + 1 && dfs(u2)) {
      matching[v] = u1;
      used[u1] = true;
      return true;
    }
  }
  return false;
}
inline int maxMatching() {
  fill(used, used + n1, false);
  fill(matching, matching + n2, -1);
  for (int res = 0;;) {
    bfs();
    fill(vis, vis + n1, false);
    int f = 0;
    for (int u = 0; u < n1; ++u)
      if (!used[u] && dfs(u)) ++f;
    if (!f) return res;
    res += f;
  }
}
int V, E, n, k;
int sp[N][N];
int par[N][N];
int home[N];
vector<pair<int, int> > adj[N];
inline void dijkstra(int s) {
  for (int i = 0; i <= V; i++) {
    sp[s][i] = INF;
    par[s][i] = -1;
  }
  sp[s][s] = 0;
  par[s][s] = -1;
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      pq;
  pq.push(pair<int, int>(0, s));
  while (!pq.empty()) {
    int u = pq.top().second;
    int d = pq.top().first;
    pq.pop();
    for (int i = 0; i < adj[u].size(); i++) {
      int v = adj[u][i].first;
      int w = adj[u][i].second;
      if (d + w < sp[s][v]) {
        sp[s][v] = d + w;
        par[s][v] = u;
        pq.push(pair<int, int>(sp[s][v], v));
      }
    }
  }
}
inline bool solve(int x) {
  init(V + 1, V + 1);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= V; j++) {
      if (sp[home[i]][j] <= x) {
        addEdge(i, j);
      }
    }
  }
  int result = maxMatching();
  return (result >= k);
}
int main() {
  scanf("%d %d %d %d", &V, &E, &n, &k);
  set<int> roots;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &home[i]);
    roots.insert(home[i]);
  }
  for (int i = 1; i <= E; i++) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    adj[u].push_back(pair<int, int>(v, w));
    adj[v].push_back(pair<int, int>(u, w));
  }
  for (int root : roots) {
    dijkstra(root);
  }
  int l = 0, r = 1731312;
  int ans = -1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (solve(mid)) {
      ans = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  printf("%d\n", ans);
}

### Prompt

Please formulate 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 N = 700;
const int S = 200;
const int M = N * S + 100;
const int INF = 1000 * 1000 * 1000;
int n1, n2, edges, last[N], previous[M], head[M];
int matching[N], dist[N], Q[N];
bool used[N], vis[N];
inline void init(int _n1, int _n2) {
  n1 = _n1;
  n2 = _n2;
  edges = 0;
  fill(last, last + n1, -1);
}
inline void addEdge(int u, int v) {
  head[edges] = v;
  previous[edges] = last[u];
  last[u] = edges++;
}
inline void bfs() {
  fill(dist, dist + n1, -1);
  int sizeQ = 0;
  for (int u = 0; u < n1; ++u) {
    if (!used[u]) {
      Q[sizeQ++] = u;
      dist[u] = 0;
    }
  }
  for (int i = 0; i < sizeQ; i++) {
    int u1 = Q[i];
    for (int e = last[u1]; e >= 0; e = previous[e]) {
      int u2 = matching[head[e]];
      if (u2 >= 0 && dist[u2] < 0) {
        dist[u2] = dist[u1] + 1;
        Q[sizeQ++] = u2;
      }
    }
  }
}
inline bool dfs(int u1) {
  vis[u1] = true;
  for (int e = last[u1]; e >= 0; e = previous[e]) {
    int v = head[e];
    int u2 = matching[v];
    if (u2 < 0 || !vis[u2] && dist[u2] == dist[u1] + 1 && dfs(u2)) {
      matching[v] = u1;
      used[u1] = true;
      return true;
    }
  }
  return false;
}
inline int maxMatching() {
  fill(used, used + n1, false);
  fill(matching, matching + n2, -1);
  for (int res = 0;;) {
    bfs();
    fill(vis, vis + n1, false);
    int f = 0;
    for (int u = 0; u < n1; ++u)
      if (!used[u] && dfs(u)) ++f;
    if (!f) return res;
    res += f;
  }
}
int V, E, n, k;
int sp[N][N];
int par[N][N];
int home[N];
vector<pair<int, int> > adj[N];
inline void dijkstra(int s) {
  for (int i = 0; i <= V; i++) {
    sp[s][i] = INF;
    par[s][i] = -1;
  }
  sp[s][s] = 0;
  par[s][s] = -1;
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      pq;
  pq.push(pair<int, int>(0, s));
  while (!pq.empty()) {
    int u = pq.top().second;
    int d = pq.top().first;
    pq.pop();
    for (int i = 0; i < adj[u].size(); i++) {
      int v = adj[u][i].first;
      int w = adj[u][i].second;
      if (d + w < sp[s][v]) {
        sp[s][v] = d + w;
        par[s][v] = u;
        pq.push(pair<int, int>(sp[s][v], v));
      }
    }
  }
}
inline bool solve(int x) {
  init(V + 1, V + 1);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= V; j++) {
      if (sp[home[i]][j] <= x) {
        addEdge(i, j);
      }
    }
  }
  int result = maxMatching();
  return (result >= k);
}
int main() {
  scanf("%d %d %d %d", &V, &E, &n, &k);
  set<int> roots;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &home[i]);
    roots.insert(home[i]);
  }
  for (int i = 1; i <= E; i++) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    adj[u].push_back(pair<int, int>(v, w));
    adj[v].push_back(pair<int, int>(u, w));
  }
  for (int root : roots) {
    dijkstra(root);
  }
  int l = 0, r = 1731312;
  int ans = -1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (solve(mid)) {
      ans = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  printf("%d\n", ans);
}

```