#include <bits/stdc++.h>
using namespace std;
const int maxn = 601;
const int INF = 1e9 + 10;
queue<int> q;
int v, e, n, k, a, b, t;
int st[maxn];
bool vis[maxn];
int dist[maxn][maxn];
vector<pair<int, int> > path[maxn];
vector<int> match[maxn];
int ptr[maxn];
bool dfs(int p) {
  for (int i = 0; i < match[p].size(); i++)
    if (!vis[match[p][i]]) {
      vis[match[p][i]] = true;
      if (ptr[match[p][i]] == 0 || dfs(ptr[match[p][i]])) {
        ptr[match[p][i]] = p;
        return true;
      }
    }
  return false;
}
bool judge(int p) {
  for (int i = 1; i <= n; i++) {
    match[i].clear();
    for (int j = 1; j <= v; j++)
      if (dist[i][j] <= p) match[i].push_back(j);
  }
  for (int i = 1; i <= v; i++) ptr[i] = 0;
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    memset(vis, false, sizeof(vis));
    if (dfs(i)) cnt++;
  }
  return cnt >= k;
}
int main() {
  scanf("%d%d%d%d", &v, &e, &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &st[i]);
  for (int i = 1; i <= e; i++) {
    scanf("%d%d%d", &a, &b, &t);
    path[a].push_back(make_pair(b, t));
    path[b].push_back(make_pair(a, t));
  }
  for (int i = 1; i <= n; i++) {
    q.push(st[i]);
    for (int j = 1; j <= v; j++) dist[i][j] = INF;
    dist[i][st[i]] = 0;
    vis[st[i]] = true;
    while (!q.empty()) {
      int cur = q.front();
      q.pop();
      vis[cur] = false;
      for (int j = 0; j < path[cur].size(); j++)
        if (dist[i][path[cur][j].first] > dist[i][cur] + path[cur][j].second) {
          dist[i][path[cur][j].first] = dist[i][cur] + path[cur][j].second;
          if (!vis[path[cur][j].first]) {
            q.push(path[cur][j].first);
            vis[path[cur][j].first] = true;
          }
        }
    }
  }
  int l = -1, r = 1731312;
  if (!judge(r)) {
    printf("-1\n");
    return 0;
  }
  while (l + 1 < r) {
    int mid = (l + r) >> 1;
    if (judge(mid))
      r = mid;
    else
      l = mid;
  }
  printf("%d\n", r);
  return 0;
}

### Prompt

Your challenge is to write 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 maxn = 601;
const int INF = 1e9 + 10;
queue<int> q;
int v, e, n, k, a, b, t;
int st[maxn];
bool vis[maxn];
int dist[maxn][maxn];
vector<pair<int, int> > path[maxn];
vector<int> match[maxn];
int ptr[maxn];
bool dfs(int p) {
  for (int i = 0; i < match[p].size(); i++)
    if (!vis[match[p][i]]) {
      vis[match[p][i]] = true;
      if (ptr[match[p][i]] == 0 || dfs(ptr[match[p][i]])) {
        ptr[match[p][i]] = p;
        return true;
      }
    }
  return false;
}
bool judge(int p) {
  for (int i = 1; i <= n; i++) {
    match[i].clear();
    for (int j = 1; j <= v; j++)
      if (dist[i][j] <= p) match[i].push_back(j);
  }
  for (int i = 1; i <= v; i++) ptr[i] = 0;
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    memset(vis, false, sizeof(vis));
    if (dfs(i)) cnt++;
  }
  return cnt >= k;
}
int main() {
  scanf("%d%d%d%d", &v, &e, &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &st[i]);
  for (int i = 1; i <= e; i++) {
    scanf("%d%d%d", &a, &b, &t);
    path[a].push_back(make_pair(b, t));
    path[b].push_back(make_pair(a, t));
  }
  for (int i = 1; i <= n; i++) {
    q.push(st[i]);
    for (int j = 1; j <= v; j++) dist[i][j] = INF;
    dist[i][st[i]] = 0;
    vis[st[i]] = true;
    while (!q.empty()) {
      int cur = q.front();
      q.pop();
      vis[cur] = false;
      for (int j = 0; j < path[cur].size(); j++)
        if (dist[i][path[cur][j].first] > dist[i][cur] + path[cur][j].second) {
          dist[i][path[cur][j].first] = dist[i][cur] + path[cur][j].second;
          if (!vis[path[cur][j].first]) {
            q.push(path[cur][j].first);
            vis[path[cur][j].first] = true;
          }
        }
    }
  }
  int l = -1, r = 1731312;
  if (!judge(r)) {
    printf("-1\n");
    return 0;
  }
  while (l + 1 < r) {
    int mid = (l + r) >> 1;
    if (judge(mid))
      r = mid;
    else
      l = mid;
  }
  printf("%d\n", r);
  return 0;
}

```