#include <bits/stdc++.h>
using namespace std;
const int maxn = 1010;
int p[maxn];
int dis[maxn][maxn];
void floyd(int v) {
  for (int i = 0; i <= v; i++) dis[i][i] = 0;
  for (int k = 1; k <= v; k++) {
    for (int i = 1; i <= v; i++) {
      for (int j = 1; j <= v; j++) {
        dis[i][j] = min(dis[i][k] + dis[k][j], dis[i][j]);
      }
    }
  }
}
int pre[maxn], vis[maxn];
int g[maxn][maxn];
int dfs(int u, int v) {
  for (int i = 1; i <= v; i++) {
    if (g[u][i] && !vis[i]) {
      vis[i] = 1;
      if (pre[i] == -1 || dfs(pre[i], v)) {
        pre[i] = u;
        return 1;
      }
    }
  }
  return 0;
}
int hungry(int n, int v) {
  int ans = 0;
  memset(pre, -1, sizeof pre);
  for (int i = 0; i < n; i++) {
    memset(vis, 0, sizeof(vis));
    if (dfs(i, v)) ans++;
  }
  return ans;
}
int main() {
  memset(dis, 0x3f, sizeof dis);
  int v, e, n, m;
  scanf("%d%d%d%d", &v, &e, &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &p[i]);
  for (int i = 0; i < e; i++) {
    int a, b, w;
    scanf("%d%d%d", &a, &b, &w);
    dis[a][b] = dis[b][a] = min(dis[b][a], w);
  }
  floyd(v);
  int l = 0, r = 2e6;
  while (l < r) {
    int mid = (l + r) / 2;
    memset(g, 0, sizeof g);
    for (int i = 0; i < n; i++) {
      for (int j = 1; j <= v; j++) {
        if (dis[p[i]][j] <= mid) g[i][j] = 1;
      }
    }
    int rec = hungry(n, v);
    if (rec >= m)
      r = mid;
    else
      l = mid + 1;
  }
  cout << (r == 2e6 ? -1 : r) << endl;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 = 1010;
int p[maxn];
int dis[maxn][maxn];
void floyd(int v) {
  for (int i = 0; i <= v; i++) dis[i][i] = 0;
  for (int k = 1; k <= v; k++) {
    for (int i = 1; i <= v; i++) {
      for (int j = 1; j <= v; j++) {
        dis[i][j] = min(dis[i][k] + dis[k][j], dis[i][j]);
      }
    }
  }
}
int pre[maxn], vis[maxn];
int g[maxn][maxn];
int dfs(int u, int v) {
  for (int i = 1; i <= v; i++) {
    if (g[u][i] && !vis[i]) {
      vis[i] = 1;
      if (pre[i] == -1 || dfs(pre[i], v)) {
        pre[i] = u;
        return 1;
      }
    }
  }
  return 0;
}
int hungry(int n, int v) {
  int ans = 0;
  memset(pre, -1, sizeof pre);
  for (int i = 0; i < n; i++) {
    memset(vis, 0, sizeof(vis));
    if (dfs(i, v)) ans++;
  }
  return ans;
}
int main() {
  memset(dis, 0x3f, sizeof dis);
  int v, e, n, m;
  scanf("%d%d%d%d", &v, &e, &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &p[i]);
  for (int i = 0; i < e; i++) {
    int a, b, w;
    scanf("%d%d%d", &a, &b, &w);
    dis[a][b] = dis[b][a] = min(dis[b][a], w);
  }
  floyd(v);
  int l = 0, r = 2e6;
  while (l < r) {
    int mid = (l + r) / 2;
    memset(g, 0, sizeof g);
    for (int i = 0; i < n; i++) {
      for (int j = 1; j <= v; j++) {
        if (dis[p[i]][j] <= mid) g[i][j] = 1;
      }
    }
    int rec = hungry(n, v);
    if (rec >= m)
      r = mid;
    else
      l = mid + 1;
  }
  cout << (r == 2e6 ? -1 : r) << endl;
}

```