#include <bits/stdc++.h>
using namespace std;
const int V = 610;
const int M = 1e9 + 7;
const int UP = 1731311;
const int MAX = 1e9;
int n, e, v, k;
int dp[V][V], w[V][V], link[V], vis[V], cur[V];
int dfs(int x) {
  for (int i = 1; i <= n; i++) {
    if (w[x][i] && !vis[i]) {
      vis[i] = 1;
      if (!link[i] || dfs(link[i])) {
        link[i] = x;
        return 1;
      }
    }
  }
  return 0;
}
int check(int ans = 0) {
  memset(link, 0, sizeof(link));
  for (int i = 1; i <= v; i++) {
    memset(vis, 0, sizeof(vis));
    ans += dfs(i);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i, j, h, a, b, c;
  cin >> n >> e >> v >> k;
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= n; j++) {
      if (i != j)
        dp[i][j] = MAX;
      else
        dp[i][j] = 0;
    }
  }
  for (i = 1; i <= v; i++) {
    cin >> cur[i];
  }
  for (i = 0; i < e; i++) {
    cin >> a >> b >> c;
    dp[a][b] = dp[b][a] = min(dp[a][b], c);
  }
  for (h = 1; h <= n; h++) {
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        if (dp[i][j] > dp[i][h] + dp[h][j]) dp[i][j] = dp[i][h] + dp[h][j];
      }
    }
  }
  int ans = -1, l = 0, r = UP;
  while (l <= r) {
    int mid = (l + r) / 2;
    for (i = 1; i <= v; i++) {
      for (j = 1; j <= n; j++) {
        w[i][j] = (dp[cur[i]][j] <= mid ? 1 : 0);
      }
    }
    if (check() >= k) {
      ans = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write 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 V = 610;
const int M = 1e9 + 7;
const int UP = 1731311;
const int MAX = 1e9;
int n, e, v, k;
int dp[V][V], w[V][V], link[V], vis[V], cur[V];
int dfs(int x) {
  for (int i = 1; i <= n; i++) {
    if (w[x][i] && !vis[i]) {
      vis[i] = 1;
      if (!link[i] || dfs(link[i])) {
        link[i] = x;
        return 1;
      }
    }
  }
  return 0;
}
int check(int ans = 0) {
  memset(link, 0, sizeof(link));
  for (int i = 1; i <= v; i++) {
    memset(vis, 0, sizeof(vis));
    ans += dfs(i);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i, j, h, a, b, c;
  cin >> n >> e >> v >> k;
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= n; j++) {
      if (i != j)
        dp[i][j] = MAX;
      else
        dp[i][j] = 0;
    }
  }
  for (i = 1; i <= v; i++) {
    cin >> cur[i];
  }
  for (i = 0; i < e; i++) {
    cin >> a >> b >> c;
    dp[a][b] = dp[b][a] = min(dp[a][b], c);
  }
  for (h = 1; h <= n; h++) {
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        if (dp[i][j] > dp[i][h] + dp[h][j]) dp[i][j] = dp[i][h] + dp[h][j];
      }
    }
  }
  int ans = -1, l = 0, r = UP;
  while (l <= r) {
    int mid = (l + r) / 2;
    for (i = 1; i <= v; i++) {
      for (j = 1; j <= n; j++) {
        w[i][j] = (dp[cur[i]][j] <= mid ? 1 : 0);
      }
    }
    if (check() >= k) {
      ans = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  cout << ans << endl;
  return 0;
}

```