#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma loop - opt(on)
using namespace std;
long long int v, e, n, k, aa, bb, cc;
long long int d[602][602], a[602], g[602][602], m[602], vis[602];
void floyd() {
  for (long long int k = 1; k <= v; k++) {
    for (long long int i = 1; i <= v; i++) {
      for (long long int j = 1; j <= v; j++) {
        d[i][j] = min(d[i][k] + d[k][j], d[i][j]);
      }
    }
  }
  return;
}
bool dfs(long long int x) {
  for (long long int i = 1; i <= v; i++) {
    if (!g[x][i] || vis[i]) continue;
    vis[i] = 1;
    if (!~m[i] || dfs(m[i])) {
      m[i] = x;
      return true;
    }
  }
  return false;
}
long long int match() {
  memset(m, -1, sizeof(m));
  long long int ans = 0;
  for (long long int i = 1; i <= n; i++) {
    memset(vis, 0, sizeof(vis));
    if (dfs(i)) ans++;
  }
  return ans;
}
bool check(long long int t) {
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= v; j++) {
      g[i][j] = (d[a[i]][j] <= t);
    }
  }
  return match() >= k;
}
long long int solve() {
  long long int l = -1, r = 1731312;
  while (r - l > 1) {
    long long int mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  if (r == 1731312)
    return -1;
  else
    return r;
}
signed main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> v >> e >> n >> k;
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  for (long long int i = 1; i <= v; i++)
    for (long long int j = 1; j <= v; j++) d[i][j] = 1000000000000000;
  for (long long int i = 1; i <= v; i++) d[i][i] = 0;
  for (long long int i = 1; i <= e; i++) {
    cin >> aa >> bb >> cc;
    d[aa][bb] = min(d[aa][bb], cc), d[bb][aa] = min(d[bb][aa], cc);
  }
  floyd();
  cout << solve() << "\n";
  return 0;
}

### 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>
#pragma GCC optimize("Ofast")
#pragma loop - opt(on)
using namespace std;
long long int v, e, n, k, aa, bb, cc;
long long int d[602][602], a[602], g[602][602], m[602], vis[602];
void floyd() {
  for (long long int k = 1; k <= v; k++) {
    for (long long int i = 1; i <= v; i++) {
      for (long long int j = 1; j <= v; j++) {
        d[i][j] = min(d[i][k] + d[k][j], d[i][j]);
      }
    }
  }
  return;
}
bool dfs(long long int x) {
  for (long long int i = 1; i <= v; i++) {
    if (!g[x][i] || vis[i]) continue;
    vis[i] = 1;
    if (!~m[i] || dfs(m[i])) {
      m[i] = x;
      return true;
    }
  }
  return false;
}
long long int match() {
  memset(m, -1, sizeof(m));
  long long int ans = 0;
  for (long long int i = 1; i <= n; i++) {
    memset(vis, 0, sizeof(vis));
    if (dfs(i)) ans++;
  }
  return ans;
}
bool check(long long int t) {
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= v; j++) {
      g[i][j] = (d[a[i]][j] <= t);
    }
  }
  return match() >= k;
}
long long int solve() {
  long long int l = -1, r = 1731312;
  while (r - l > 1) {
    long long int mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  if (r == 1731312)
    return -1;
  else
    return r;
}
signed main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> v >> e >> n >> k;
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  for (long long int i = 1; i <= v; i++)
    for (long long int j = 1; j <= v; j++) d[i][j] = 1000000000000000;
  for (long long int i = 1; i <= v; i++) d[i][i] = 0;
  for (long long int i = 1; i <= e; i++) {
    cin >> aa >> bb >> cc;
    d[aa][bb] = min(d[aa][bb], cc), d[bb][aa] = min(d[bb][aa], cc);
  }
  floyd();
  cout << solve() << "\n";
  return 0;
}

```