#include <bits/stdc++.h>
using namespace std;
const int maxn = 5000;
const int maxM = (int)1e6 + 100;
const int mod = (int)1e9 + 7;
const int maxlog = (int)20;
const int P = mod;
int v, E, n, k, st[maxn], dp[maxn][maxn], mt[maxn], used[maxn];
vector<pair<int, int> > r[maxn];
vector<int> c[maxn];
vector<int> g[maxn];
int dfs(int u) {
  if (used[u]) return 0;
  used[u] = 1;
  for (int to : g[u]) {
    if (mt[to] == -1 || dfs(mt[to])) {
      mt[to] = u;
      return 1;
    }
  }
  return 0;
}
inline void build(int second) {
  for (int i = 0; i < maxn; i++) dp[second][i] = INT_MAX;
  dp[second][second] = 0;
  set<pair<int, int> > q;
  q.insert(make_pair(dp[second][second], second));
  while (!q.empty()) {
    int v = q.begin()->second;
    q.erase(q.begin());
    for (int i = 0; i < r[v].size(); i++) {
      int to = r[v][i].first, len = r[v][i].second;
      if (dp[second][v] + len < dp[second][to]) {
        q.erase(make_pair(dp[second][to], to));
        dp[second][to] = dp[second][v] + len;
        q.insert(make_pair(dp[second][to], to));
      }
    }
  }
}
inline int check(int x) {
  for (int i = 0; i < maxn; i++) g[i].clear();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= v; j++) {
      if (dp[st[i]][j] <= x) g[i].push_back(j);
    }
  }
  int ans = 0;
  memset(mt, -1, sizeof(mt));
  for (int i = 1; i <= n; i++) {
    memset(used, 0, sizeof(used));
    if (dfs(i)) ans++;
  }
  if (ans >= k) return 1;
  return 0;
}
int main() {
  cin >> v >> E >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> st[i];
  }
  for (int i = 1; i <= E; i++) {
    int u, v, t;
    cin >> u >> v >> t;
    r[u].push_back(make_pair(v, t));
    r[v].push_back(make_pair(u, t));
  }
  for (int i = 1; i <= v; i++) build(i);
  int l = 0, r = 1731311, res = -1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid)) {
      res = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  cout << res;
}

### Prompt

Create a solution in Cpp for 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 maxn = 5000;
const int maxM = (int)1e6 + 100;
const int mod = (int)1e9 + 7;
const int maxlog = (int)20;
const int P = mod;
int v, E, n, k, st[maxn], dp[maxn][maxn], mt[maxn], used[maxn];
vector<pair<int, int> > r[maxn];
vector<int> c[maxn];
vector<int> g[maxn];
int dfs(int u) {
  if (used[u]) return 0;
  used[u] = 1;
  for (int to : g[u]) {
    if (mt[to] == -1 || dfs(mt[to])) {
      mt[to] = u;
      return 1;
    }
  }
  return 0;
}
inline void build(int second) {
  for (int i = 0; i < maxn; i++) dp[second][i] = INT_MAX;
  dp[second][second] = 0;
  set<pair<int, int> > q;
  q.insert(make_pair(dp[second][second], second));
  while (!q.empty()) {
    int v = q.begin()->second;
    q.erase(q.begin());
    for (int i = 0; i < r[v].size(); i++) {
      int to = r[v][i].first, len = r[v][i].second;
      if (dp[second][v] + len < dp[second][to]) {
        q.erase(make_pair(dp[second][to], to));
        dp[second][to] = dp[second][v] + len;
        q.insert(make_pair(dp[second][to], to));
      }
    }
  }
}
inline int check(int x) {
  for (int i = 0; i < maxn; i++) g[i].clear();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= v; j++) {
      if (dp[st[i]][j] <= x) g[i].push_back(j);
    }
  }
  int ans = 0;
  memset(mt, -1, sizeof(mt));
  for (int i = 1; i <= n; i++) {
    memset(used, 0, sizeof(used));
    if (dfs(i)) ans++;
  }
  if (ans >= k) return 1;
  return 0;
}
int main() {
  cin >> v >> E >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> st[i];
  }
  for (int i = 1; i <= E; i++) {
    int u, v, t;
    cin >> u >> v >> t;
    r[u].push_back(make_pair(v, t));
    r[v].push_back(make_pair(u, t));
  }
  for (int i = 1; i <= v; i++) build(i);
  int l = 0, r = 1731311, res = -1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid)) {
      res = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  cout << res;
}

```