#include <bits/stdc++.h>
using namespace std;
const int maxn = 600 + 17;
const int maxv = 2 + 2 * maxn;
int V, E, N, K, S, T;
int init[maxn], dist[maxn][maxn];
vector<int> vx[maxv];
vector<pair<int, int> > graph[maxn];
int flow[maxv][maxv];
int prew[maxv];
bool fflow[maxv];
inline void add(int u, int v) {
  vx[u].push_back(v);
  vx[v].push_back(u);
}
bool BFS() {
  memset(fflow, 0, sizeof fflow);
  memset(prew, 0xFF, sizeof prew);
  queue<int> q;
  q.push(S);
  fflow[S] = true;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int v : vx[u])
      if (!fflow[v] && flow[u][v]) {
        prew[v] = u, fflow[v] = true, q.push(v);
        if (v == T) break;
      }
  }
  return fflow[T];
}
bool solve(int limit) {
  memset(flow, 0, sizeof flow);
  for (int i = 0; i <= T; i++) vx[i].clear();
  for (int i = 1; i <= N; i++) add(S, i), flow[S][i] = 1;
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= V; j++)
      if (dist[init[i]][j] <= limit)
        add(i, 1 + N + (j - 1)), flow[i][1 + N + (j - 1)] = 1;
  for (int i = 1; i <= V; i++)
    add(1 + N + (i - 1), T), flow[1 + N + (i - 1)][T] = 1;
  int total = 0;
  while (BFS()) {
    int u, cflow = (1 << 28);
    u = T;
    while (prew[u] >= 0) cflow = min(cflow, flow[prew[u]][u]), u = prew[u];
    u = T;
    while (prew[u] >= 0)
      flow[prew[u]][u] -= cflow, flow[u][prew[u]] += cflow, u = prew[u];
    total += cflow;
  }
  return total >= K;
}
int main() {
  ios_base::sync_with_stdio(false);
  memset(dist, 0x2F, sizeof dist);
  for (int i = 1; i < maxn; i++) dist[i][i] = 0;
  cin >> V >> E >> N >> K;
  for (int i = 1; i <= N; i++) cin >> init[i];
  for (int i = 1; i <= E; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    dist[x][y] = dist[y][x] = min(dist[x][y], z);
  }
  for (int k = 1; k <= V; k++)
    for (int i = 1; i <= V; i++)
      for (int j = 1; j <= V; j++)
        dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
  S = 0;
  T = 1 + N + V;
  int low = 0, high = (1 << 28), mid, ret = (1 << 28);
  while (low <= high) {
    mid = low + high >> 1;
    if (solve(mid))
      high = mid - 1, ret = mid;
    else
      low = mid + 1;
  }
  cout << (ret == (1 << 28) ? -1 : ret) << endl;
  return 0;
}

### Prompt

Please formulate 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 maxn = 600 + 17;
const int maxv = 2 + 2 * maxn;
int V, E, N, K, S, T;
int init[maxn], dist[maxn][maxn];
vector<int> vx[maxv];
vector<pair<int, int> > graph[maxn];
int flow[maxv][maxv];
int prew[maxv];
bool fflow[maxv];
inline void add(int u, int v) {
  vx[u].push_back(v);
  vx[v].push_back(u);
}
bool BFS() {
  memset(fflow, 0, sizeof fflow);
  memset(prew, 0xFF, sizeof prew);
  queue<int> q;
  q.push(S);
  fflow[S] = true;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int v : vx[u])
      if (!fflow[v] && flow[u][v]) {
        prew[v] = u, fflow[v] = true, q.push(v);
        if (v == T) break;
      }
  }
  return fflow[T];
}
bool solve(int limit) {
  memset(flow, 0, sizeof flow);
  for (int i = 0; i <= T; i++) vx[i].clear();
  for (int i = 1; i <= N; i++) add(S, i), flow[S][i] = 1;
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= V; j++)
      if (dist[init[i]][j] <= limit)
        add(i, 1 + N + (j - 1)), flow[i][1 + N + (j - 1)] = 1;
  for (int i = 1; i <= V; i++)
    add(1 + N + (i - 1), T), flow[1 + N + (i - 1)][T] = 1;
  int total = 0;
  while (BFS()) {
    int u, cflow = (1 << 28);
    u = T;
    while (prew[u] >= 0) cflow = min(cflow, flow[prew[u]][u]), u = prew[u];
    u = T;
    while (prew[u] >= 0)
      flow[prew[u]][u] -= cflow, flow[u][prew[u]] += cflow, u = prew[u];
    total += cflow;
  }
  return total >= K;
}
int main() {
  ios_base::sync_with_stdio(false);
  memset(dist, 0x2F, sizeof dist);
  for (int i = 1; i < maxn; i++) dist[i][i] = 0;
  cin >> V >> E >> N >> K;
  for (int i = 1; i <= N; i++) cin >> init[i];
  for (int i = 1; i <= E; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    dist[x][y] = dist[y][x] = min(dist[x][y], z);
  }
  for (int k = 1; k <= V; k++)
    for (int i = 1; i <= V; i++)
      for (int j = 1; j <= V; j++)
        dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
  S = 0;
  T = 1 + N + V;
  int low = 0, high = (1 << 28), mid, ret = (1 << 28);
  while (low <= high) {
    mid = low + high >> 1;
    if (solve(mid))
      high = mid - 1, ret = mid;
    else
      low = mid + 1;
  }
  cout << (ret == (1 << 28) ? -1 : ret) << endl;
  return 0;
}

```