#include <bits/stdc++.h>
using namespace std;
class MaxFlow {
 public:
  struct Edge {
    int b, e, f;
  } E[501000 * 2];
  vector<int> G[2000];
  int Level[2000], Q[2000], PV[2000], source, sink, n, flow, EC;
  void init(int N, int S, int T) {
    n = N, flow = EC = 0;
    source = S, sink = T;
  }
  void Add_Edge(int a, int b, int f) {
    G[a].push_back(EC);
    G[b].push_back(EC + 1);
    E[EC++] = {a, b, f};
    E[EC++] = {b, a, 0};
  }
  bool GetLevel() {
    int i;
    for (i = 1; i <= n; i++) Level[i] = -1;
    int head = 0, tail = 0;
    Q[++tail] = source, Level[source] = 0;
    while (head < tail) {
      int x = Q[++head];
      for (auto &y : G[x]) {
        if (E[y].f && Level[E[y].e] == -1) {
          Q[++tail] = E[y].e;
          Level[E[y].e] = Level[x] + 1;
        }
      }
    }
    return Level[sink] != -1;
  }
  int BlockFlow(int a, int f) {
    if (a == sink) return f;
    for (int &i = PV[a]; i >= 0; i--) {
      int x = G[a][i];
      if (E[x].f && Level[E[x].e] == Level[a] + 1) {
        int t = BlockFlow(E[x].e, min(f, E[x].f));
        if (t) {
          E[x].f -= t;
          E[x ^ 1].f += t;
          return t;
        }
      }
    }
    return 0;
  }
  void Dinic() {
    int t;
    while (GetLevel()) {
      for (int i = 1; i <= n; i++) PV[i] = G[i].size() - 1;
      while (t = BlockFlow(source, 9999999)) flow += t;
    }
  }
} G;
int n, m, K, T;
int w[2000], D[2000][2000], CK[2000];
bool Pos(int mid) {
  G.init(n + n + 2, n + n + 1, n + n + 2);
  int i, j;
  for (i = 1; i <= n; i++) CK[i] = 0;
  for (i = 1; i <= K; i++) {
    CK[w[i]]++;
  }
  for (i = 1; i <= n; i++)
    if (CK[i]) G.Add_Edge(G.source, i, CK[i]);
  for (i = 1; i <= n; i++) {
    G.Add_Edge(n + i, G.sink, 1);
    if (!CK[i]) continue;
    for (j = 1; j <= n; j++) {
      if (D[i][j] <= mid) {
        G.Add_Edge(i, n + j, 1);
      }
    }
  }
  G.Dinic();
  for (i = 1; i <= G.n; i++) G.G[i].clear();
  return G.flow >= T;
}
int main() {
  int i, j, a, b, c, k;
  scanf("%d%d%d%d", &n, &m, &K, &T);
  for (i = 1; i <= K; i++) {
    scanf("%d", &w[i]);
  }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++)
      if (i != j) D[i][j] = 1e9;
  for (i = 0; i < m; i++) {
    scanf("%d%d%d", &a, &b, &c);
    D[a][b] = min(D[a][b], c);
    D[b][a] = D[a][b];
  }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++)
      for (k = 1; k <= n; k++) D[j][k] = min(D[j][k], D[j][i] + D[i][k]);
  int be = 0, ed = 1e8, res = -1, mid;
  while (be <= ed) {
    mid = (be + ed) >> 1;
    if (Pos(mid)) {
      res = mid;
      ed = mid - 1;
    } else
      be = mid + 1;
  }
  printf("%d\n", res);
}

### Prompt

Generate 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;
class MaxFlow {
 public:
  struct Edge {
    int b, e, f;
  } E[501000 * 2];
  vector<int> G[2000];
  int Level[2000], Q[2000], PV[2000], source, sink, n, flow, EC;
  void init(int N, int S, int T) {
    n = N, flow = EC = 0;
    source = S, sink = T;
  }
  void Add_Edge(int a, int b, int f) {
    G[a].push_back(EC);
    G[b].push_back(EC + 1);
    E[EC++] = {a, b, f};
    E[EC++] = {b, a, 0};
  }
  bool GetLevel() {
    int i;
    for (i = 1; i <= n; i++) Level[i] = -1;
    int head = 0, tail = 0;
    Q[++tail] = source, Level[source] = 0;
    while (head < tail) {
      int x = Q[++head];
      for (auto &y : G[x]) {
        if (E[y].f && Level[E[y].e] == -1) {
          Q[++tail] = E[y].e;
          Level[E[y].e] = Level[x] + 1;
        }
      }
    }
    return Level[sink] != -1;
  }
  int BlockFlow(int a, int f) {
    if (a == sink) return f;
    for (int &i = PV[a]; i >= 0; i--) {
      int x = G[a][i];
      if (E[x].f && Level[E[x].e] == Level[a] + 1) {
        int t = BlockFlow(E[x].e, min(f, E[x].f));
        if (t) {
          E[x].f -= t;
          E[x ^ 1].f += t;
          return t;
        }
      }
    }
    return 0;
  }
  void Dinic() {
    int t;
    while (GetLevel()) {
      for (int i = 1; i <= n; i++) PV[i] = G[i].size() - 1;
      while (t = BlockFlow(source, 9999999)) flow += t;
    }
  }
} G;
int n, m, K, T;
int w[2000], D[2000][2000], CK[2000];
bool Pos(int mid) {
  G.init(n + n + 2, n + n + 1, n + n + 2);
  int i, j;
  for (i = 1; i <= n; i++) CK[i] = 0;
  for (i = 1; i <= K; i++) {
    CK[w[i]]++;
  }
  for (i = 1; i <= n; i++)
    if (CK[i]) G.Add_Edge(G.source, i, CK[i]);
  for (i = 1; i <= n; i++) {
    G.Add_Edge(n + i, G.sink, 1);
    if (!CK[i]) continue;
    for (j = 1; j <= n; j++) {
      if (D[i][j] <= mid) {
        G.Add_Edge(i, n + j, 1);
      }
    }
  }
  G.Dinic();
  for (i = 1; i <= G.n; i++) G.G[i].clear();
  return G.flow >= T;
}
int main() {
  int i, j, a, b, c, k;
  scanf("%d%d%d%d", &n, &m, &K, &T);
  for (i = 1; i <= K; i++) {
    scanf("%d", &w[i]);
  }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++)
      if (i != j) D[i][j] = 1e9;
  for (i = 0; i < m; i++) {
    scanf("%d%d%d", &a, &b, &c);
    D[a][b] = min(D[a][b], c);
    D[b][a] = D[a][b];
  }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++)
      for (k = 1; k <= n; k++) D[j][k] = min(D[j][k], D[j][i] + D[i][k]);
  int be = 0, ed = 1e8, res = -1, mid;
  while (be <= ed) {
    mid = (be + ed) >> 1;
    if (Pos(mid)) {
      res = mid;
      ed = mid - 1;
    } else
      be = mid + 1;
  }
  printf("%d\n", res);
}

```