#include <bits/stdc++.h>
using namespace std;
struct Node {
  int d, u;
  bool operator<(const Node& rhs) const { return d > rhs.d; }
};
vector<Node> c[610];
int d[610][610], g[610][610];
int mat[1210], s[610];
vector<int> f[1210];
bool used[1210];
bool vis[610];
priority_queue<Node> q;
bool crosspath(int x) {
  for (int i = 0; i < f[x].size(); i++) {
    int j = f[x][i];
    if (!used[j]) {
      used[j] = 1;
      if (mat[j] == 0 || crosspath(mat[j])) {
        mat[j] = x;
        return 1;
      }
    }
  }
  return 0;
}
int main() {
  int v, e, n, kk, i, j, a, b, t;
  scanf("%d%d%d%d", &v, &e, &n, &kk);
  for (i = 1; i <= n; i++) scanf("%d", &s[i]);
  for (i = 1; i <= v; i++)
    for (j = 1; j <= v; j++) g[i][j] = 10001;
  for (i = 1; i <= e; i++) {
    scanf("%d%d%d", &a, &b, &t);
    if (g[a][b] > t) {
      g[a][b] = t;
      g[b][a] = t;
    }
  }
  for (i = 1; i < v; i++)
    for (j = i + 1; j <= v; j++) {
      if (g[i][j] <= 10000) {
        t = g[i][j];
        c[i].push_back((Node){t, j});
        c[j].push_back((Node){t, i});
      }
    }
  memset(d, 70, sizeof(d));
  for (i = 1; i <= n; i++) {
    d[i][s[i]] = 0;
    memset(vis, 0, sizeof(vis));
    while (!q.empty()) q.pop();
    q.push((Node){0, s[i]});
    while (!q.empty()) {
      Node x = q.top();
      q.pop();
      int u = x.u;
      if (vis[u]) continue;
      vis[u] = 1;
      for (int k = 0; k < c[u].size(); k++) {
        int v = c[u][k].u;
        int w = c[u][k].d;
        if (d[i][v] > d[i][u] + w) {
          d[i][v] = d[i][u] + w;
          q.push((Node){d[i][v], v});
        }
      }
    }
  }
  int l = 0, r = 1731311, ans = r + 1;
  while (l <= r) {
    int mid = (l + r) / 2;
    for (i = 1; i <= n; i++) f[i].clear();
    for (i = 1; i <= n; i++) {
      for (int j = 1; j <= v; j++)
        if (d[i][j] <= mid) f[i].push_back(j + v);
    }
    int match = 0;
    memset(mat, 0, sizeof(mat));
    for (i = 1; i <= n; i++) {
      memset(used, 0, sizeof(used));
      if (crosspath(i)) match++;
    }
    if (match >= kk) {
      if (ans > mid) ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  if (ans > 1731311)
    printf("-1\n");
  else
    printf("%d\n", ans);
}

### Prompt

Please create a solution in Cpp 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;
struct Node {
  int d, u;
  bool operator<(const Node& rhs) const { return d > rhs.d; }
};
vector<Node> c[610];
int d[610][610], g[610][610];
int mat[1210], s[610];
vector<int> f[1210];
bool used[1210];
bool vis[610];
priority_queue<Node> q;
bool crosspath(int x) {
  for (int i = 0; i < f[x].size(); i++) {
    int j = f[x][i];
    if (!used[j]) {
      used[j] = 1;
      if (mat[j] == 0 || crosspath(mat[j])) {
        mat[j] = x;
        return 1;
      }
    }
  }
  return 0;
}
int main() {
  int v, e, n, kk, i, j, a, b, t;
  scanf("%d%d%d%d", &v, &e, &n, &kk);
  for (i = 1; i <= n; i++) scanf("%d", &s[i]);
  for (i = 1; i <= v; i++)
    for (j = 1; j <= v; j++) g[i][j] = 10001;
  for (i = 1; i <= e; i++) {
    scanf("%d%d%d", &a, &b, &t);
    if (g[a][b] > t) {
      g[a][b] = t;
      g[b][a] = t;
    }
  }
  for (i = 1; i < v; i++)
    for (j = i + 1; j <= v; j++) {
      if (g[i][j] <= 10000) {
        t = g[i][j];
        c[i].push_back((Node){t, j});
        c[j].push_back((Node){t, i});
      }
    }
  memset(d, 70, sizeof(d));
  for (i = 1; i <= n; i++) {
    d[i][s[i]] = 0;
    memset(vis, 0, sizeof(vis));
    while (!q.empty()) q.pop();
    q.push((Node){0, s[i]});
    while (!q.empty()) {
      Node x = q.top();
      q.pop();
      int u = x.u;
      if (vis[u]) continue;
      vis[u] = 1;
      for (int k = 0; k < c[u].size(); k++) {
        int v = c[u][k].u;
        int w = c[u][k].d;
        if (d[i][v] > d[i][u] + w) {
          d[i][v] = d[i][u] + w;
          q.push((Node){d[i][v], v});
        }
      }
    }
  }
  int l = 0, r = 1731311, ans = r + 1;
  while (l <= r) {
    int mid = (l + r) / 2;
    for (i = 1; i <= n; i++) f[i].clear();
    for (i = 1; i <= n; i++) {
      for (int j = 1; j <= v; j++)
        if (d[i][j] <= mid) f[i].push_back(j + v);
    }
    int match = 0;
    memset(mat, 0, sizeof(mat));
    for (i = 1; i <= n; i++) {
      memset(used, 0, sizeof(used));
      if (crosspath(i)) match++;
    }
    if (match >= kk) {
      if (ans > mid) ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  if (ans > 1731311)
    printf("-1\n");
  else
    printf("%d\n", ans);
}

```