#include <bits/stdc++.h>
using namespace std;
int V, E, N, K, ren[205], link[605];
int mp[605][605], dst[605], mid;
bool vis[605], used[605];
vector<int> v[605];
bool dfs(int a) {
  for (int i = 1; i <= V; i++) {
    if (mp[a][i] != -1 && !vis[i] && mp[a][i] <= mid) {
      vis[i] = 1;
      if (link[i] == -1 || dfs(link[i])) {
        link[i] = a;
        return 1;
      }
    }
  }
  return 0;
}
bool solve() {
  int ans = 0;
  memset(link, -1, sizeof(link));
  for (int i = 1; i <= N; i++) {
    memset(vis, 0, sizeof(vis));
    ans += dfs(ren[i]);
  }
  if (ans >= K) return 1;
  return 0;
}
void SPFA(int a) {
  queue<int> q;
  memset(dst, -1, sizeof(dst));
  memset(vis, 0, sizeof(vis));
  dst[a] = 0;
  q.push(a);
  while (!q.empty()) {
    int p = q.front();
    q.pop();
    vis[p] = 0;
    for (int i = 0; i < v[p].size(); i++) {
      int u = v[p][i];
      if (dst[u] == -1 || dst[u] > dst[p] + mp[p][u]) {
        dst[u] = dst[p] + mp[p][u];
        if (!vis[u]) {
          q.push(u);
          vis[u] = 1;
        }
      }
    }
  }
  for (int i = 1; i <= V; i++) mp[a][i] = dst[i];
}
int main() {
  int i, j, a, b, c;
  while (~scanf("%d %d %d %d", &V, &E, &N, &K)) {
    memset(mp, -1, sizeof(mp));
    memset(used, 0, sizeof(used));
    for (i = 1; i <= N; i++) scanf("%d", ren + i);
    for (i = 1; i <= V; i++) {
      mp[i][i] = 0;
      v[i].clear();
    }
    while (E--) {
      scanf("%d %d %d", &a, &b, &c);
      if (mp[a][b] == -1 || mp[a][b] > c) {
        mp[a][b] = mp[b][a] = c;
        v[a].push_back(b);
        v[b].push_back(a);
      }
    }
    for (i = 1; i <= N; i++) {
      if (!used[ren[i]]) {
        SPFA(ren[i]);
        used[ren[i]] = 1;
      }
    }
    int r = 1731311;
    int l = 0;
    while (l <= r) {
      mid = (r + l) / 2;
      if (solve())
        r = mid - 1;
      else
        l = mid + 1;
    }
    if (l > 1731311)
      printf("-1\n");
    else
      printf("%d\n", l);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 V, E, N, K, ren[205], link[605];
int mp[605][605], dst[605], mid;
bool vis[605], used[605];
vector<int> v[605];
bool dfs(int a) {
  for (int i = 1; i <= V; i++) {
    if (mp[a][i] != -1 && !vis[i] && mp[a][i] <= mid) {
      vis[i] = 1;
      if (link[i] == -1 || dfs(link[i])) {
        link[i] = a;
        return 1;
      }
    }
  }
  return 0;
}
bool solve() {
  int ans = 0;
  memset(link, -1, sizeof(link));
  for (int i = 1; i <= N; i++) {
    memset(vis, 0, sizeof(vis));
    ans += dfs(ren[i]);
  }
  if (ans >= K) return 1;
  return 0;
}
void SPFA(int a) {
  queue<int> q;
  memset(dst, -1, sizeof(dst));
  memset(vis, 0, sizeof(vis));
  dst[a] = 0;
  q.push(a);
  while (!q.empty()) {
    int p = q.front();
    q.pop();
    vis[p] = 0;
    for (int i = 0; i < v[p].size(); i++) {
      int u = v[p][i];
      if (dst[u] == -1 || dst[u] > dst[p] + mp[p][u]) {
        dst[u] = dst[p] + mp[p][u];
        if (!vis[u]) {
          q.push(u);
          vis[u] = 1;
        }
      }
    }
  }
  for (int i = 1; i <= V; i++) mp[a][i] = dst[i];
}
int main() {
  int i, j, a, b, c;
  while (~scanf("%d %d %d %d", &V, &E, &N, &K)) {
    memset(mp, -1, sizeof(mp));
    memset(used, 0, sizeof(used));
    for (i = 1; i <= N; i++) scanf("%d", ren + i);
    for (i = 1; i <= V; i++) {
      mp[i][i] = 0;
      v[i].clear();
    }
    while (E--) {
      scanf("%d %d %d", &a, &b, &c);
      if (mp[a][b] == -1 || mp[a][b] > c) {
        mp[a][b] = mp[b][a] = c;
        v[a].push_back(b);
        v[b].push_back(a);
      }
    }
    for (i = 1; i <= N; i++) {
      if (!used[ren[i]]) {
        SPFA(ren[i]);
        used[ren[i]] = 1;
      }
    }
    int r = 1731311;
    int l = 0;
    while (l <= r) {
      mid = (r + l) / 2;
      if (solve())
        r = mid - 1;
      else
        l = mid + 1;
    }
    if (l > 1731311)
      printf("-1\n");
    else
      printf("%d\n", l);
  }
  return 0;
}

```