#include <bits/stdc++.h>
using namespace std;
const int N = 605;
int V, e, n, K, x[N], dis[N][N];
int vis[N], mch[N];
vector<int> G[N];
int dfs(int u, int s) {
  if (vis[u] == s) return 0;
  vis[u] = s;
  for (int v : G[u])
    if (!mch[v] || dfs(mch[v], s)) return mch[v] = u, 1;
  return 0;
}
int chk(int mid, int res = 0) {
  for (int i = (1); i <= (n); i++) {
    G[i].clear();
    for (int j = (1); j <= (V); j++)
      if (dis[x[i]][j] <= mid) G[i].push_back(j);
  }
  memset(vis, 0, sizeof vis), memset(mch, 0, sizeof mch);
  for (int i = (1); i <= (n); i++)
    if ((res += dfs(i, i)) == K) return 1;
  return 0;
}
int main() {
  memset(dis, 63, sizeof dis);
  cin >> V >> e >> n >> K;
  for (int i = (1); i <= (n); i++) scanf("%d", &x[i]);
  for (int i = (1); i <= (V); i++) dis[i][i] = 0;
  for (int u, v, w; e--; 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 <= (V); k++)
    for (int i = (1); i <= (V); i++)
      for (int j = (1); j <= (V); j++)
        (dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]));
  if (!chk(0x3f3f3f3f - 1)) puts("-1"), exit(0);
  int l = 0, r = V * 10000, mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (chk(mid))
      r = mid - 1;
    else
      l = mid + 1;
  }
  cout << 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;
const int N = 605;
int V, e, n, K, x[N], dis[N][N];
int vis[N], mch[N];
vector<int> G[N];
int dfs(int u, int s) {
  if (vis[u] == s) return 0;
  vis[u] = s;
  for (int v : G[u])
    if (!mch[v] || dfs(mch[v], s)) return mch[v] = u, 1;
  return 0;
}
int chk(int mid, int res = 0) {
  for (int i = (1); i <= (n); i++) {
    G[i].clear();
    for (int j = (1); j <= (V); j++)
      if (dis[x[i]][j] <= mid) G[i].push_back(j);
  }
  memset(vis, 0, sizeof vis), memset(mch, 0, sizeof mch);
  for (int i = (1); i <= (n); i++)
    if ((res += dfs(i, i)) == K) return 1;
  return 0;
}
int main() {
  memset(dis, 63, sizeof dis);
  cin >> V >> e >> n >> K;
  for (int i = (1); i <= (n); i++) scanf("%d", &x[i]);
  for (int i = (1); i <= (V); i++) dis[i][i] = 0;
  for (int u, v, w; e--; 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 <= (V); k++)
    for (int i = (1); i <= (V); i++)
      for (int j = (1); j <= (V); j++)
        (dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]));
  if (!chk(0x3f3f3f3f - 1)) puts("-1"), exit(0);
  int l = 0, r = V * 10000, mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (chk(mid))
      r = mid - 1;
    else
      l = mid + 1;
  }
  cout << l;
  return 0;
}

```