#include <bits/stdc++.h>
using namespace std;
const int N = 600 + 5;
int n, e, m, k, dis[N][N], start[201], vis[201], match[N], lol;
vector<vector<pair<int, int>>> v;
vector<vector<int>> temp;
priority_queue<pair<int, int>> q;
void dijk(int u) {
  q.push({0, u});
  dis[u][u] = 0;
  while (!q.empty()) {
    pair<int, int> x = q.top();
    x.first *= -1;
    q.pop();
    if (dis[u][x.second] < x.first) continue;
    for (int i = 0; i < v[x.second].size(); i++) {
      int a = v[x.second][i].first, b = v[x.second][i].second;
      if (dis[u][a] <= x.first + b) continue;
      dis[u][a] = x.first + b;
      q.push({-dis[u][a], a});
    }
  }
}
bool dfs(int u) {
  vis[u] = lol;
  for (int i = 0; i < temp[u].size(); i++) {
    int a = temp[u][i];
    if (match[a] == -1 || vis[match[a]] != lol && dfs(match[a])) {
      match[a] = u;
      return true;
    }
  }
  return false;
}
bool check(int mid) {
  temp.clear();
  temp.resize(m);
  memset(match, -1, sizeof match);
  for (int i = 0; i < m; i++)
    for (int j = 0; j < n; j++)
      if (dis[start[i]][j] <= mid) temp[i].push_back(j);
  int ret = 0;
  for (int i = 0; i < m; i++, lol++) ret += dfs(i);
  return ret >= k;
}
int main() {
  scanf("%d%d%d%d", &n, &e, &m, &k);
  for (int i = 0; i < m; i++) scanf("%d", start + i), start[i]--;
  v.resize(n);
  for (int i = 0; i < e; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    v[--a].push_back({--b, c});
    v[b].push_back({a, c});
  }
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++) dis[i][j] = 1e9;
  for (int i = 0; i < n; i++) dijk(i);
  int l = 0, r = 1731311;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (check(mid))
      r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%d\n", l > 1731311 ? -1 : l);
}

### 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 N = 600 + 5;
int n, e, m, k, dis[N][N], start[201], vis[201], match[N], lol;
vector<vector<pair<int, int>>> v;
vector<vector<int>> temp;
priority_queue<pair<int, int>> q;
void dijk(int u) {
  q.push({0, u});
  dis[u][u] = 0;
  while (!q.empty()) {
    pair<int, int> x = q.top();
    x.first *= -1;
    q.pop();
    if (dis[u][x.second] < x.first) continue;
    for (int i = 0; i < v[x.second].size(); i++) {
      int a = v[x.second][i].first, b = v[x.second][i].second;
      if (dis[u][a] <= x.first + b) continue;
      dis[u][a] = x.first + b;
      q.push({-dis[u][a], a});
    }
  }
}
bool dfs(int u) {
  vis[u] = lol;
  for (int i = 0; i < temp[u].size(); i++) {
    int a = temp[u][i];
    if (match[a] == -1 || vis[match[a]] != lol && dfs(match[a])) {
      match[a] = u;
      return true;
    }
  }
  return false;
}
bool check(int mid) {
  temp.clear();
  temp.resize(m);
  memset(match, -1, sizeof match);
  for (int i = 0; i < m; i++)
    for (int j = 0; j < n; j++)
      if (dis[start[i]][j] <= mid) temp[i].push_back(j);
  int ret = 0;
  for (int i = 0; i < m; i++, lol++) ret += dfs(i);
  return ret >= k;
}
int main() {
  scanf("%d%d%d%d", &n, &e, &m, &k);
  for (int i = 0; i < m; i++) scanf("%d", start + i), start[i]--;
  v.resize(n);
  for (int i = 0; i < e; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    v[--a].push_back({--b, c});
    v[b].push_back({a, c});
  }
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++) dis[i][j] = 1e9;
  for (int i = 0; i < n; i++) dijk(i);
  int l = 0, r = 1731311;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (check(mid))
      r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%d\n", l > 1731311 ? -1 : l);
}

```