#include <bits/stdc++.h>
using namespace std;
template <class T>
class flow_graph {
 public:
  static constexpr T eps = (T)1e-9;
  struct edge {
    int to;
    T c;
    T f;
    int rev;
  };
  vector<vector<edge> > g;
  vector<int> ptr;
  vector<int> d;
  int n;
  int st, fin;
  bool modified;
  flow_graph(int n, int st, int fin) : n(n), st(st), fin(fin) {
    modified = true;
    assert(0 <= st && st < n && 0 <= fin && fin < n && st != fin);
    g.resize(n);
    ptr.resize(n);
    d.resize(n);
  }
  void clear_flow() {
    modified = true;
    for (int i = 0; i < n; i++) {
      for (edge &e : g[i]) {
        e.f = 0;
      }
    }
  }
  void add(int from, int to, T forward_cap, T backward_cap) {
    modified = true;
    assert(0 <= from && from < n && 0 <= to && to < n);
    int from_size = g[from].size();
    int to_size = g[to].size();
    g[from].push_back({to, forward_cap, 0, to_size});
    g[to].push_back({from, backward_cap, 0, from_size});
  }
  bool expath() {
    queue<int> q({st});
    fill(d.begin(), d.end(), -1);
    d[st] = 0;
    while (!q.empty()) {
      int i = q.front();
      q.pop();
      for (edge &e : g[i]) {
        if (e.c > e.f + eps && d[e.to] == -1) {
          d[e.to] = d[i] + 1;
          if (e.to == fin) {
            return true;
          }
          q.push(e.to);
        }
      }
    }
    return false;
  }
  T dfs(int v, T w) {
    if (v == fin) {
      return w;
    }
    T r = 0;
    int j = ptr[v];
    while (j > 0) {
      j--;
      edge &e = g[v][j];
      if (e.c > e.f + eps && d[e.to] == d[v] + 1) {
        T ww = e.c - e.f;
        if (v != st) {
          ww = min(ww, w - r);
        }
        T t = dfs(e.to, ww);
        if (t > 0) {
          e.f += t;
          g[e.to][e.rev].f -= t;
          r += t;
          if (v != st && r >= w - eps) {
            return r;
          }
        }
      }
      ptr[v]--;
    }
    return r;
  }
  T flow;
  T max_flow() {
    if (!modified) {
      return flow;
    }
    modified = false;
    T ans = 0;
    while (expath()) {
      for (int i = 0; i < n; i++) {
        ptr[i] = g[i].size();
      }
      T add = dfs(st, -1);
      if (add <= eps) {
        break;
      }
      ans += add;
    }
    return flow = ans;
  }
  vector<bool> min_cut() {
    max_flow();
    vector<bool> ret(n);
    for (int i = 0; i < n; i++) {
      ret[i] = (d[i] != -1);
    }
    return ret;
  }
};
const int N = 1234;
int city[N];
int g[N][N];
int main() {
  int v, e, n, k;
  scanf("%d %d %d %d", &v, &e, &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", city + i);
    city[i]--;
  }
  for (int i = 0; i < v; i++) {
    for (int j = 0; j < v; j++) {
      g[i][j] = (i == j ? 0 : (int)1e9);
    }
  }
  for (int i = 0; i < e; i++) {
    int x, y, z;
    scanf("%d %d %d", &x, &y, &z);
    x--;
    y--;
    g[x][y] = min(g[x][y], z);
    g[y][x] = min(g[y][x], z);
  }
  for (int k = 0; k < v; k++) {
    for (int i = 0; i < v; i++) {
      for (int j = 0; j < v; j++) {
        g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
      }
    }
  }
  const int inf = (int)2e6;
  int low = 0, high = inf;
  while (low < high) {
    int mid = (low + high) >> 1;
    flow_graph<int> gr(n + v + 2, n + v, n + v + 1);
    for (int i = 0; i < n; i++) {
      gr.add(n + v, i, 1, 0);
    }
    for (int j = 0; j < v; j++) {
      gr.add(n + j, n + v + 1, 1, 0);
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < v; j++) {
        if (g[city[i]][j] <= mid) {
          gr.add(i, n + j, 1, 0);
        }
      }
    }
    if (gr.max_flow() >= k) {
      high = mid;
    } else {
      low = mid + 1;
    }
  }
  printf("%d\n", low == inf ? -1 : low);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
template <class T>
class flow_graph {
 public:
  static constexpr T eps = (T)1e-9;
  struct edge {
    int to;
    T c;
    T f;
    int rev;
  };
  vector<vector<edge> > g;
  vector<int> ptr;
  vector<int> d;
  int n;
  int st, fin;
  bool modified;
  flow_graph(int n, int st, int fin) : n(n), st(st), fin(fin) {
    modified = true;
    assert(0 <= st && st < n && 0 <= fin && fin < n && st != fin);
    g.resize(n);
    ptr.resize(n);
    d.resize(n);
  }
  void clear_flow() {
    modified = true;
    for (int i = 0; i < n; i++) {
      for (edge &e : g[i]) {
        e.f = 0;
      }
    }
  }
  void add(int from, int to, T forward_cap, T backward_cap) {
    modified = true;
    assert(0 <= from && from < n && 0 <= to && to < n);
    int from_size = g[from].size();
    int to_size = g[to].size();
    g[from].push_back({to, forward_cap, 0, to_size});
    g[to].push_back({from, backward_cap, 0, from_size});
  }
  bool expath() {
    queue<int> q({st});
    fill(d.begin(), d.end(), -1);
    d[st] = 0;
    while (!q.empty()) {
      int i = q.front();
      q.pop();
      for (edge &e : g[i]) {
        if (e.c > e.f + eps && d[e.to] == -1) {
          d[e.to] = d[i] + 1;
          if (e.to == fin) {
            return true;
          }
          q.push(e.to);
        }
      }
    }
    return false;
  }
  T dfs(int v, T w) {
    if (v == fin) {
      return w;
    }
    T r = 0;
    int j = ptr[v];
    while (j > 0) {
      j--;
      edge &e = g[v][j];
      if (e.c > e.f + eps && d[e.to] == d[v] + 1) {
        T ww = e.c - e.f;
        if (v != st) {
          ww = min(ww, w - r);
        }
        T t = dfs(e.to, ww);
        if (t > 0) {
          e.f += t;
          g[e.to][e.rev].f -= t;
          r += t;
          if (v != st && r >= w - eps) {
            return r;
          }
        }
      }
      ptr[v]--;
    }
    return r;
  }
  T flow;
  T max_flow() {
    if (!modified) {
      return flow;
    }
    modified = false;
    T ans = 0;
    while (expath()) {
      for (int i = 0; i < n; i++) {
        ptr[i] = g[i].size();
      }
      T add = dfs(st, -1);
      if (add <= eps) {
        break;
      }
      ans += add;
    }
    return flow = ans;
  }
  vector<bool> min_cut() {
    max_flow();
    vector<bool> ret(n);
    for (int i = 0; i < n; i++) {
      ret[i] = (d[i] != -1);
    }
    return ret;
  }
};
const int N = 1234;
int city[N];
int g[N][N];
int main() {
  int v, e, n, k;
  scanf("%d %d %d %d", &v, &e, &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", city + i);
    city[i]--;
  }
  for (int i = 0; i < v; i++) {
    for (int j = 0; j < v; j++) {
      g[i][j] = (i == j ? 0 : (int)1e9);
    }
  }
  for (int i = 0; i < e; i++) {
    int x, y, z;
    scanf("%d %d %d", &x, &y, &z);
    x--;
    y--;
    g[x][y] = min(g[x][y], z);
    g[y][x] = min(g[y][x], z);
  }
  for (int k = 0; k < v; k++) {
    for (int i = 0; i < v; i++) {
      for (int j = 0; j < v; j++) {
        g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
      }
    }
  }
  const int inf = (int)2e6;
  int low = 0, high = inf;
  while (low < high) {
    int mid = (low + high) >> 1;
    flow_graph<int> gr(n + v + 2, n + v, n + v + 1);
    for (int i = 0; i < n; i++) {
      gr.add(n + v, i, 1, 0);
    }
    for (int j = 0; j < v; j++) {
      gr.add(n + j, n + v + 1, 1, 0);
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < v; j++) {
        if (g[city[i]][j] <= mid) {
          gr.add(i, n + j, 1, 0);
        }
      }
    }
    if (gr.max_flow() >= k) {
      high = mid;
    } else {
      low = mid + 1;
    }
  }
  printf("%d\n", low == inf ? -1 : low);
  return 0;
}

```