#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 30;
const long long LINF = 1ll << 60;
const int MOD = 1000000007;
const int SIZE_ = 610;
vector<vector<int>> g(SIZE_), dist(SIZE_);
vector<vector<pair<int, int>>> gc(SIZE_);
vector<int> mt(SIZE_, -1), mat(SIZE_);
vector<int> uA(SIZE_);
set<int> ans;
vector<int> c(SIZE_);
int n, m, k, cnt;
bool kun(int v) {
  if (uA[v]) return false;
  uA[v] = true;
  for (auto& i : g[v]) {
    if (mt[i] == -1) {
      mt[i] = v;
      return true;
    }
  }
  for (auto& i : g[v]) {
    if (kun(mt[i])) {
      mt[i] = v;
      return true;
    }
  }
  return false;
}
int starts[SIZE_];
void dej(int v, vector<int>& d) {
  d.resize(m, INF);
  d[v] = 0;
  set<pair<int, int>> st;
  st.insert({0, v});
  while (st.size()) {
    int v = st.begin()->second;
    st.erase(st.begin());
    for (auto& i : gc[v]) {
      if (d[i.first] > d[v] + i.second) {
        st.erase({d[i.first], i.first});
        d[i.first] = d[v] + i.second;
        st.insert({d[i.first], i.first});
      }
    }
  }
}
bool check(int t) {
  for (int i = 0; i < n; ++i) {
    g[i].clear();
    for (int j = 0; j < m; ++j) {
      if (dist[i][j] <= t) g[i].push_back(j);
    }
  }
  mt.assign(SIZE_, -1);
  mat.assign(SIZE_, 0);
  int k = 0;
  while (1) {
    int rs = 0;
    uA.assign(SIZE_, 0);
    for (int i = 0; i < n; ++i) {
      if (!mat[i]) {
        if (kun(i)) {
          ++rs;
          mat[i] = 1;
        }
      }
    }
    if (!rs) break;
    k += rs;
  }
  return k >= cnt;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> m >> k >> n >> cnt;
  for (int i = 0; i < n; ++i) {
    cin >> starts[i];
    --starts[i];
  }
  for (int i = 0; i < k; ++i) {
    int x, y, v;
    cin >> x >> y >> v;
    gc[y - 1].push_back({x - 1, v});
    gc[x - 1].push_back({y - 1, v});
  }
  for (int i = 0; i < n; ++i) {
    dej(starts[i], dist[i]);
  }
  int l = 0, r = 1731311;
  while (l + 1 < r) {
    int b = (l + r) / 2;
    if (check(b))
      r = b;
    else
      l = b;
  }
  if (check(l))
    cout << l;
  else if (check(r))
    cout << r;
  else
    cout << -1;
  return 0;
}

### Prompt

Your challenge is to write 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 INF = 1 << 30;
const long long LINF = 1ll << 60;
const int MOD = 1000000007;
const int SIZE_ = 610;
vector<vector<int>> g(SIZE_), dist(SIZE_);
vector<vector<pair<int, int>>> gc(SIZE_);
vector<int> mt(SIZE_, -1), mat(SIZE_);
vector<int> uA(SIZE_);
set<int> ans;
vector<int> c(SIZE_);
int n, m, k, cnt;
bool kun(int v) {
  if (uA[v]) return false;
  uA[v] = true;
  for (auto& i : g[v]) {
    if (mt[i] == -1) {
      mt[i] = v;
      return true;
    }
  }
  for (auto& i : g[v]) {
    if (kun(mt[i])) {
      mt[i] = v;
      return true;
    }
  }
  return false;
}
int starts[SIZE_];
void dej(int v, vector<int>& d) {
  d.resize(m, INF);
  d[v] = 0;
  set<pair<int, int>> st;
  st.insert({0, v});
  while (st.size()) {
    int v = st.begin()->second;
    st.erase(st.begin());
    for (auto& i : gc[v]) {
      if (d[i.first] > d[v] + i.second) {
        st.erase({d[i.first], i.first});
        d[i.first] = d[v] + i.second;
        st.insert({d[i.first], i.first});
      }
    }
  }
}
bool check(int t) {
  for (int i = 0; i < n; ++i) {
    g[i].clear();
    for (int j = 0; j < m; ++j) {
      if (dist[i][j] <= t) g[i].push_back(j);
    }
  }
  mt.assign(SIZE_, -1);
  mat.assign(SIZE_, 0);
  int k = 0;
  while (1) {
    int rs = 0;
    uA.assign(SIZE_, 0);
    for (int i = 0; i < n; ++i) {
      if (!mat[i]) {
        if (kun(i)) {
          ++rs;
          mat[i] = 1;
        }
      }
    }
    if (!rs) break;
    k += rs;
  }
  return k >= cnt;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> m >> k >> n >> cnt;
  for (int i = 0; i < n; ++i) {
    cin >> starts[i];
    --starts[i];
  }
  for (int i = 0; i < k; ++i) {
    int x, y, v;
    cin >> x >> y >> v;
    gc[y - 1].push_back({x - 1, v});
    gc[x - 1].push_back({y - 1, v});
  }
  for (int i = 0; i < n; ++i) {
    dej(starts[i], dist[i]);
  }
  int l = 0, r = 1731311;
  while (l + 1 < r) {
    int b = (l + r) / 2;
    if (check(b))
      r = b;
    else
      l = b;
  }
  if (check(l))
    cout << l;
  else if (check(r))
    cout << r;
  else
    cout << -1;
  return 0;
}

```