#include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 10;
const int M = 1000000007;
const double PI = atan(1) * 4;
const int oo = 1000000000;
int n, m, t, k, path[601][601], pos[201], vs, match[601], vis[201];
vector<vector<pair<int, int> > > g;
vector<vector<int> > w;
bool get(int u) {
  vis[u] = vs;
  for (int i = 0; i < w[u].size(); ++i) {
    int d = w[u][i];
    if (match[d] == -1 || vis[match[d]] != vs && get(match[d])) {
      match[d] = u;
      return true;
    }
  }
  return false;
}
bool can(int tm) {
  w.clear();
  w.resize(n);
  for (int i = 0; i < t; ++i)
    for (int j = 0; j < n; ++j)
      if (path[pos[i]][j] <= tm) w[i].push_back(j);
  int res = 0;
  memset(match, -1, sizeof(match));
  for (int i = 0; i < t; ++i) {
    ++vs;
    res += get(i);
  }
  return res >= k;
}
void dijk(int u) {
  priority_queue<pair<int, int> > q;
  q.push(pair<int, int>(0, u));
  while (!q.empty()) {
    int d = q.top().second, cst = -q.top().first;
    q.pop();
    if (cst > path[u][d]) continue;
    for (int i = 0; i < g[d].size(); ++i) {
      int x = g[d][i].first, cs = g[d][i].second;
      if (path[u][d] + cs < path[u][x]) {
        path[u][x] = path[u][d] + cs;
        q.push(pair<int, int>(-path[u][x], x));
      }
    }
  }
}
int main() {
  cin >> n >> m >> t >> k;
  g.resize(n);
  for (int i = 0; i < t; ++i) {
    scanf("%d", &pos[i]);
    --pos[i];
  }
  for (int a, b, c, i = 0; i < m; ++i) {
    scanf("%d%d%d", &a, &b, &c);
    --a;
    --b;
    g[a].push_back(pair<int, int>(b, c));
    g[b].push_back(pair<int, int>(a, c));
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      if (i != j) path[i][j] = oo;
  for (int i = 0; i < n; ++i) dijk(i);
  int lo = 0, hi = 1731311, md, bst = oo;
  while (lo <= hi) {
    md = (lo + hi) / 2;
    if (can(md)) {
      bst = md;
      hi = md - 1;
    } else
      lo = md + 1;
  }
  if (bst == oo)
    cout << -1 << endl;
  else
    cout << bst << endl;
}

### Prompt

Your task is to create 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 N = 100000 + 10;
const int M = 1000000007;
const double PI = atan(1) * 4;
const int oo = 1000000000;
int n, m, t, k, path[601][601], pos[201], vs, match[601], vis[201];
vector<vector<pair<int, int> > > g;
vector<vector<int> > w;
bool get(int u) {
  vis[u] = vs;
  for (int i = 0; i < w[u].size(); ++i) {
    int d = w[u][i];
    if (match[d] == -1 || vis[match[d]] != vs && get(match[d])) {
      match[d] = u;
      return true;
    }
  }
  return false;
}
bool can(int tm) {
  w.clear();
  w.resize(n);
  for (int i = 0; i < t; ++i)
    for (int j = 0; j < n; ++j)
      if (path[pos[i]][j] <= tm) w[i].push_back(j);
  int res = 0;
  memset(match, -1, sizeof(match));
  for (int i = 0; i < t; ++i) {
    ++vs;
    res += get(i);
  }
  return res >= k;
}
void dijk(int u) {
  priority_queue<pair<int, int> > q;
  q.push(pair<int, int>(0, u));
  while (!q.empty()) {
    int d = q.top().second, cst = -q.top().first;
    q.pop();
    if (cst > path[u][d]) continue;
    for (int i = 0; i < g[d].size(); ++i) {
      int x = g[d][i].first, cs = g[d][i].second;
      if (path[u][d] + cs < path[u][x]) {
        path[u][x] = path[u][d] + cs;
        q.push(pair<int, int>(-path[u][x], x));
      }
    }
  }
}
int main() {
  cin >> n >> m >> t >> k;
  g.resize(n);
  for (int i = 0; i < t; ++i) {
    scanf("%d", &pos[i]);
    --pos[i];
  }
  for (int a, b, c, i = 0; i < m; ++i) {
    scanf("%d%d%d", &a, &b, &c);
    --a;
    --b;
    g[a].push_back(pair<int, int>(b, c));
    g[b].push_back(pair<int, int>(a, c));
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      if (i != j) path[i][j] = oo;
  for (int i = 0; i < n; ++i) dijk(i);
  int lo = 0, hi = 1731311, md, bst = oo;
  while (lo <= hi) {
    md = (lo + hi) / 2;
    if (can(md)) {
      bst = md;
      hi = md - 1;
    } else
      lo = md + 1;
  }
  if (bst == oo)
    cout << -1 << endl;
  else
    cout << bst << endl;
}

```