#include <bits/stdc++.h>
using namespace std;
const int N = 607;
const int M = 20003;
const int base = 1e9 + 7;
const int logmax = 24;
const long long mod = 1e9 + 7;
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
long long Rand(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(rng);
}
int n, m, p, k, a[203], d[N][N], sum, f[203][N], pre[N];
vector<pair<int, int> > adj[N];
bool kt[N];
void input() {
  cin >> n >> m >> p >> k;
  for (int i = 1; i <= p; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) d[i][j] = base;
  for (int i = 1; i <= n; i++) d[i][i] = 0;
  int u, v, c;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v >> c;
    d[u][v] = d[v][u] = min(d[u][v], c);
  }
}
void init() {
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}
int dfs(int u) {
  for (int i = 1; i <= n; i++) {
    if (!f[u][i] || kt[i]) continue;
    kt[i] = 1;
    if (!pre[i] || dfs(pre[i])) {
      pre[i] = u;
      return 1;
    }
  }
  return 0;
}
int change() {
  memset(pre, 0, sizeof(pre));
  int res = 0;
  for (int i = 1; i <= p; i++) {
    memset(kt, 0, sizeof(kt));
    if (dfs(i)) res++;
  }
  return res;
}
void sub2() {
  int l = 0, r = 2000000, res = -1;
  while (l <= r) {
    int mid = (l + r) / 2;
    memset(f, 0, sizeof(f));
    for (int i = 1; i <= p; i++)
      for (int j = 1; j <= n; j++) f[i][j] = (d[a[i]][j] <= mid ? 1 : 0);
    if (change() >= k) {
      res = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << res;
}
void sol() { sub2(); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  input();
  init();
  sol();
  return 0;
}

### Prompt

Generate 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 = 607;
const int M = 20003;
const int base = 1e9 + 7;
const int logmax = 24;
const long long mod = 1e9 + 7;
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
long long Rand(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(rng);
}
int n, m, p, k, a[203], d[N][N], sum, f[203][N], pre[N];
vector<pair<int, int> > adj[N];
bool kt[N];
void input() {
  cin >> n >> m >> p >> k;
  for (int i = 1; i <= p; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) d[i][j] = base;
  for (int i = 1; i <= n; i++) d[i][i] = 0;
  int u, v, c;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v >> c;
    d[u][v] = d[v][u] = min(d[u][v], c);
  }
}
void init() {
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}
int dfs(int u) {
  for (int i = 1; i <= n; i++) {
    if (!f[u][i] || kt[i]) continue;
    kt[i] = 1;
    if (!pre[i] || dfs(pre[i])) {
      pre[i] = u;
      return 1;
    }
  }
  return 0;
}
int change() {
  memset(pre, 0, sizeof(pre));
  int res = 0;
  for (int i = 1; i <= p; i++) {
    memset(kt, 0, sizeof(kt));
    if (dfs(i)) res++;
  }
  return res;
}
void sub2() {
  int l = 0, r = 2000000, res = -1;
  while (l <= r) {
    int mid = (l + r) / 2;
    memset(f, 0, sizeof(f));
    for (int i = 1; i <= p; i++)
      for (int j = 1; j <= n; j++) f[i][j] = (d[a[i]][j] <= mid ? 1 : 0);
    if (change() >= k) {
      res = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << res;
}
void sol() { sub2(); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  input();
  init();
  sol();
  return 0;
}

```