#include <bits/stdc++.h>
using namespace std;
const int N = 201;
const int maxn = 601;
int n, m, nn;
vector<int> cites[maxn];
int mp[N][maxn], link[maxn], k, used[maxn], cnt, Mp[maxn][maxn], city[maxn];
void floyd() {
  for (int h = 1; h <= n; h++)
    for (int j = 1; j <= n; j++)
      for (int i = 1; i <= n; i++)
        if (i != j && i != h && j != h)
          Mp[j][i] = min(Mp[j][i], Mp[j][h] + Mp[h][i]);
}
bool dfs(int u, int Max) {
  for (int i = 1; i <= n; i++) {
    if (mp[u][i] <= Max && !used[i]) {
      used[i] = 1;
      if (link[i] == -1 || dfs(link[i], Max)) {
        link[i] = u;
        return true;
      }
    }
  }
  return false;
}
int Hungary(int Max) {
  int ans = 0;
  memset(link, -1, sizeof(link));
  for (int i = 1; i <= nn; i++) {
    memset(used, 0, sizeof(used));
    if (dfs(i, Max)) ans++;
  }
  if (ans < k) return 0;
  return 1;
}
int main() {
  scanf("%d %d %d %d", &n, &m, &nn, &k);
  for (int i = 0; i <= n; i++) cites[i].clear();
  cnt = 0;
  int u, v, w;
  for (int i = 1; i <= nn; i++) {
    scanf("%d", &u);
    if (!cites[u].size()) city[cnt++] = u;
    cites[u].push_back(i);
  }
  memset(Mp, 0x3f, sizeof(Mp));
  memset(mp, 0x3f, sizeof(mp));
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &u, &v, &w);
    Mp[u][v] = min(Mp[u][v], w);
    Mp[v][u] = Mp[u][v];
  }
  if (k < cnt) {
    puts("0");
    return 0;
  }
  floyd();
  for (int i = 0; i < cnt; i++) {
    u = city[i];
    for (int j = 0; j < cites[u].size(); j++) {
      int v = cites[u][j];
      for (int h = 1; h <= n; h++) {
        if (h == u)
          mp[v][u] = 0;
        else if (Mp[u][h] != 0x3f3f3f3f)
          mp[v][h] = Mp[u][h];
      }
    }
  }
  int l = 0, r = 1731311, mid;
  while (l <= r) {
    mid = (l + r) / 2;
    int ans = Hungary(mid);
    if (ans)
      r = mid - 1;
    else
      l = mid + 1;
  }
  if (l > 1731311)
    puts("-1");
  else
    printf("%d\n", l);
  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;
const int N = 201;
const int maxn = 601;
int n, m, nn;
vector<int> cites[maxn];
int mp[N][maxn], link[maxn], k, used[maxn], cnt, Mp[maxn][maxn], city[maxn];
void floyd() {
  for (int h = 1; h <= n; h++)
    for (int j = 1; j <= n; j++)
      for (int i = 1; i <= n; i++)
        if (i != j && i != h && j != h)
          Mp[j][i] = min(Mp[j][i], Mp[j][h] + Mp[h][i]);
}
bool dfs(int u, int Max) {
  for (int i = 1; i <= n; i++) {
    if (mp[u][i] <= Max && !used[i]) {
      used[i] = 1;
      if (link[i] == -1 || dfs(link[i], Max)) {
        link[i] = u;
        return true;
      }
    }
  }
  return false;
}
int Hungary(int Max) {
  int ans = 0;
  memset(link, -1, sizeof(link));
  for (int i = 1; i <= nn; i++) {
    memset(used, 0, sizeof(used));
    if (dfs(i, Max)) ans++;
  }
  if (ans < k) return 0;
  return 1;
}
int main() {
  scanf("%d %d %d %d", &n, &m, &nn, &k);
  for (int i = 0; i <= n; i++) cites[i].clear();
  cnt = 0;
  int u, v, w;
  for (int i = 1; i <= nn; i++) {
    scanf("%d", &u);
    if (!cites[u].size()) city[cnt++] = u;
    cites[u].push_back(i);
  }
  memset(Mp, 0x3f, sizeof(Mp));
  memset(mp, 0x3f, sizeof(mp));
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &u, &v, &w);
    Mp[u][v] = min(Mp[u][v], w);
    Mp[v][u] = Mp[u][v];
  }
  if (k < cnt) {
    puts("0");
    return 0;
  }
  floyd();
  for (int i = 0; i < cnt; i++) {
    u = city[i];
    for (int j = 0; j < cites[u].size(); j++) {
      int v = cites[u][j];
      for (int h = 1; h <= n; h++) {
        if (h == u)
          mp[v][u] = 0;
        else if (Mp[u][h] != 0x3f3f3f3f)
          mp[v][h] = Mp[u][h];
      }
    }
  }
  int l = 0, r = 1731311, mid;
  while (l <= r) {
    mid = (l + r) / 2;
    int ans = Hungary(mid);
    if (ans)
      r = mid - 1;
    else
      l = mid + 1;
  }
  if (l > 1731311)
    puts("-1");
  else
    printf("%d\n", l);
  return 0;
}

```