#include <bits/stdc++.h>
using namespace std;
const int V = 601;
const int N = 201;
int cost[V][V], g[N][V], l[V], nodes, teams;
int start[N];
bool seen[V];
bool kuhn(int u, int thresh) {
  for (int v = 1; v <= nodes; ++v) {
    if (!seen[v] && g[u][v] <= thresh) {
      seen[v] = 1;
      if (l[v] == -1 || kuhn(l[v], thresh)) {
        l[v] = u;
        return true;
      }
    }
  }
  return false;
}
int maxMatch(int thresh) {
  int res = 0;
  fill(l + 1, l + nodes + 1, -1);
  for (int i = 1; i <= teams; ++i) {
    fill(seen + 1, seen + nodes + 1, false);
    res += kuhn(i, thresh);
  }
  return res;
}
int bsearch(int L, int R, int K) {
  int mid;
  while (R - L > 1) {
    mid = (L + R) / 2;
    (maxMatch(mid) >= K ? R : L) = mid;
  }
  return R;
}
int main() {
  int v, e, n, k, u, t, a, b;
  scanf("%d %d %d %d", &v, &e, &n, &k);
  nodes = v;
  teams = n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", start + i);
  }
  for (int i = 1; i <= v; ++i) {
    for (int j = 1; j <= v; ++j) {
      cost[i][j] = 1e9;
    }
    cost[i][i] = 0;
  }
  while (e--) {
    scanf("%d %d %d", &a, &b, &t);
    cost[b][a] = cost[a][b] = min(cost[a][b], t);
  }
  for (int w = 1; w <= v; w++) {
    for (int i = 1; i <= v; ++i) {
      for (int j = 1; j <= v; ++j) {
        cost[i][j] = min(cost[i][j], cost[i][w] + cost[w][j]);
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= v; ++j) {
      g[i][j] = cost[start[i]][j];
    }
  }
  int maxAns = 1731311 + 1;
  int res = bsearch(-1, maxAns, k);
  if (res >= maxAns) res = -1;
  printf("%d", res);
  return 0;
}

### Prompt

Create a solution in cpp for 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 = 601;
const int N = 201;
int cost[V][V], g[N][V], l[V], nodes, teams;
int start[N];
bool seen[V];
bool kuhn(int u, int thresh) {
  for (int v = 1; v <= nodes; ++v) {
    if (!seen[v] && g[u][v] <= thresh) {
      seen[v] = 1;
      if (l[v] == -1 || kuhn(l[v], thresh)) {
        l[v] = u;
        return true;
      }
    }
  }
  return false;
}
int maxMatch(int thresh) {
  int res = 0;
  fill(l + 1, l + nodes + 1, -1);
  for (int i = 1; i <= teams; ++i) {
    fill(seen + 1, seen + nodes + 1, false);
    res += kuhn(i, thresh);
  }
  return res;
}
int bsearch(int L, int R, int K) {
  int mid;
  while (R - L > 1) {
    mid = (L + R) / 2;
    (maxMatch(mid) >= K ? R : L) = mid;
  }
  return R;
}
int main() {
  int v, e, n, k, u, t, a, b;
  scanf("%d %d %d %d", &v, &e, &n, &k);
  nodes = v;
  teams = n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", start + i);
  }
  for (int i = 1; i <= v; ++i) {
    for (int j = 1; j <= v; ++j) {
      cost[i][j] = 1e9;
    }
    cost[i][i] = 0;
  }
  while (e--) {
    scanf("%d %d %d", &a, &b, &t);
    cost[b][a] = cost[a][b] = min(cost[a][b], t);
  }
  for (int w = 1; w <= v; w++) {
    for (int i = 1; i <= v; ++i) {
      for (int j = 1; j <= v; ++j) {
        cost[i][j] = min(cost[i][j], cost[i][w] + cost[w][j]);
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= v; ++j) {
      g[i][j] = cost[start[i]][j];
    }
  }
  int maxAns = 1731311 + 1;
  int res = bsearch(-1, maxAns, k);
  if (res >= maxAns) res = -1;
  printf("%d", res);
  return 0;
}

```