#include <bits/stdc++.h>
using namespace std;
int dis[666][666], n, m, t, l, r, p, mid, g[666][666];
int linker[666], a[666];
bool v[666];
bool dfs(int u, int x) {
  for (int i = 1; i <= n; i++)
    if (dis[a[u]][i] <= x && !v[i]) {
      v[i] = true;
      if (linker[i] == -1 || dfs(linker[i], x)) {
        linker[i] = u;
        return true;
      }
    }
  return false;
}
bool check(int x) {
  memset(linker, -1, sizeof(linker));
  int tmp = 0;
  for (int i = 1; i <= p; i++) {
    memset(v, 0, sizeof(v));
    if (dfs(i, x)) tmp++;
  }
  return tmp >= t;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &p, &t);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i == j)
        dis[i][j] = 0;
      else
        dis[i][j] = 1000000007;
  for (int i = 1; i <= p; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    dis[u][v] = min(dis[u][v], w);
    dis[v][u] = min(dis[v][u], w);
  }
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        if (dis[i][k] + dis[k][j] < dis[i][j])
          dis[i][j] = dis[i][k] + dis[k][j];
  l = 0;
  r = 1731311;
  while (r - l > 1) {
    mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  if (check(l))
    printf("%d\n", l);
  else if (!check(r))
    printf("-1\n");
  else
    printf("%d\n", r);
  return 0;
}

### Prompt

Generate 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;
int dis[666][666], n, m, t, l, r, p, mid, g[666][666];
int linker[666], a[666];
bool v[666];
bool dfs(int u, int x) {
  for (int i = 1; i <= n; i++)
    if (dis[a[u]][i] <= x && !v[i]) {
      v[i] = true;
      if (linker[i] == -1 || dfs(linker[i], x)) {
        linker[i] = u;
        return true;
      }
    }
  return false;
}
bool check(int x) {
  memset(linker, -1, sizeof(linker));
  int tmp = 0;
  for (int i = 1; i <= p; i++) {
    memset(v, 0, sizeof(v));
    if (dfs(i, x)) tmp++;
  }
  return tmp >= t;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &p, &t);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i == j)
        dis[i][j] = 0;
      else
        dis[i][j] = 1000000007;
  for (int i = 1; i <= p; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    dis[u][v] = min(dis[u][v], w);
    dis[v][u] = min(dis[v][u], w);
  }
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        if (dis[i][k] + dis[k][j] < dis[i][j])
          dis[i][j] = dis[i][k] + dis[k][j];
  l = 0;
  r = 1731311;
  while (r - l > 1) {
    mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  if (check(l))
    printf("%d\n", l);
  else if (!check(r))
    printf("-1\n");
  else
    printf("%d\n", r);
  return 0;
}

```