#include <bits/stdc++.h>
using namespace std;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const int maxn = 601;
int d[maxn][maxn];
int st[maxn], vis[maxn], match[maxn];
vector<int> w[maxn];
bool dfs(int v) {
  if (vis[v]) return false;
  vis[v] = 1;
  for (int u : w[v]) {
    if (match[u] == -1 || dfs(match[u])) {
      match[u] = v;
      return true;
    }
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  int v, e, n, k;
  cin >> v >> e >> n >> k;
  for (int i = 0; i < (int)(v); ++i)
    for (int j = 0; j < (int)(v); ++j) d[i][j] = 1e9;
  for (int i = 0; i < (int)(v); ++i) d[i][i] = 0;
  for (int i = 0; i < (int)(n); ++i) cin >> st[i], --st[i];
  for (int i = 0; i < (int)(e); ++i) {
    int x, y, t;
    cin >> x >> y >> t;
    --x;
    --y;
    uin(d[x][y], t);
    uin(d[y][x], t);
  }
  for (int k = 0; k < (int)(v); ++k)
    for (int i = 0; i < (int)(v); ++i)
      for (int j = 0; j < (int)(v); ++j) uin(d[i][j], d[i][k] + d[k][j]);
  int L = -1, R = 1e9;
  while (R - L > 1) {
    int M = (L + R) / 2;
    for (int i = 0; i < (int)(n); ++i) w[i].clear();
    for (int i = 0; i < (int)(n); ++i)
      for (int j = 0; j < (int)(v); ++j)
        if (d[st[i]][j] <= M) w[i].push_back(j);
    for (int j = 0; j < (int)(v); ++j) match[j] = -1;
    int res = 0;
    for (int i = 0; i < (int)(n); ++i) {
      for (int j = 0; j < (int)(n); ++j) vis[j] = 0;
      res += int(dfs(i));
    }
    (res >= k ? R : L) = M;
  }
  if (R > 1e8) R = -1;
  cout << R << '\n';
  return 0;
}

### 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;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const int maxn = 601;
int d[maxn][maxn];
int st[maxn], vis[maxn], match[maxn];
vector<int> w[maxn];
bool dfs(int v) {
  if (vis[v]) return false;
  vis[v] = 1;
  for (int u : w[v]) {
    if (match[u] == -1 || dfs(match[u])) {
      match[u] = v;
      return true;
    }
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  int v, e, n, k;
  cin >> v >> e >> n >> k;
  for (int i = 0; i < (int)(v); ++i)
    for (int j = 0; j < (int)(v); ++j) d[i][j] = 1e9;
  for (int i = 0; i < (int)(v); ++i) d[i][i] = 0;
  for (int i = 0; i < (int)(n); ++i) cin >> st[i], --st[i];
  for (int i = 0; i < (int)(e); ++i) {
    int x, y, t;
    cin >> x >> y >> t;
    --x;
    --y;
    uin(d[x][y], t);
    uin(d[y][x], t);
  }
  for (int k = 0; k < (int)(v); ++k)
    for (int i = 0; i < (int)(v); ++i)
      for (int j = 0; j < (int)(v); ++j) uin(d[i][j], d[i][k] + d[k][j]);
  int L = -1, R = 1e9;
  while (R - L > 1) {
    int M = (L + R) / 2;
    for (int i = 0; i < (int)(n); ++i) w[i].clear();
    for (int i = 0; i < (int)(n); ++i)
      for (int j = 0; j < (int)(v); ++j)
        if (d[st[i]][j] <= M) w[i].push_back(j);
    for (int j = 0; j < (int)(v); ++j) match[j] = -1;
    int res = 0;
    for (int i = 0; i < (int)(n); ++i) {
      for (int j = 0; j < (int)(n); ++j) vis[j] = 0;
      res += int(dfs(i));
    }
    (res >= k ? R : L) = M;
  }
  if (R > 1e8) R = -1;
  cout << R << '\n';
  return 0;
}

```