#include <bits/stdc++.h>
using namespace std;
template <typename T>
T sqr(const T &x) {
  return x * x;
}
template <typename T>
T my_abs(const T &x) {
  return x < 0 ? -x : x;
}
const int INF = (int)1e9;
const long long LINF = (long long)4e18;
const long long INF64 = LINF;
const long long INFLL = LINF;
const long double EPS = 1e-9;
const long double PI = 3.1415926535897932384626433832795;
const int N = 600 + 9;
const int M = 109;
const int MOD = int(1e9) + 7;
int n, m, k, need;
int st[N];
int d[N][N];
vector<int> g[N];
int mt[N];
bool u[N];
bool read() {
  if (!(cin >> n >> m >> k >> need)) return false;
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(n); ++j) d[i][j] = INF;
  for (int i = 0; i < int(n); ++i) d[i][i] = 0;
  for (int i = 0; i < int(k); ++i) cin >> st[i], --st[i];
  for (int i = 0; i < int(m); ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    --u, --v;
    d[u][v] = min(d[u][v], w);
    d[v][u] = min(d[v][u], w);
  }
  return true;
}
bool dfs(int v) {
  if (u[v]) return false;
  u[v] = true;
  for (int i = 0; i < int(int(g[v].size())); ++i) {
    int to = g[v][i];
    if (mt[to] == -1 || dfs(mt[to])) {
      mt[to] = v;
      return true;
    }
  }
  return false;
}
bool ok(int maxD) {
  for (int i = 0; i < int(n); ++i) g[i].clear();
  for (int i = 0; i < int(k); ++i)
    for (int j = 0; j < int(n); ++j)
      if (d[st[i]][j] <= maxD) g[i].push_back(j);
  memset(mt, -1, sizeof mt);
  int res = 0;
  for (int i = 0; i < int(k); ++i) {
    memset(u, 0, sizeof u);
    res += dfs(i);
  }
  return res >= need;
}
void solve() {
  for (int k = 0; k < int(n); ++k)
    for (int i = 0; i < int(n); ++i)
      for (int j = 0; j < int(n); ++j)
        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
  int l = -1, r = 1731311 + 100;
  while (r - l > 1) {
    int mid = (l + r) / 2;
    if (ok(mid))
      r = mid;
    else
      l = mid;
  }
  if (!ok(r))
    puts("-1");
  else
    cout << r << endl;
}
int main() {
  srand((unsigned int)time(NULL));
  cout << setprecision(15) << fixed;
  read();
  solve();
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 <typename T>
T sqr(const T &x) {
  return x * x;
}
template <typename T>
T my_abs(const T &x) {
  return x < 0 ? -x : x;
}
const int INF = (int)1e9;
const long long LINF = (long long)4e18;
const long long INF64 = LINF;
const long long INFLL = LINF;
const long double EPS = 1e-9;
const long double PI = 3.1415926535897932384626433832795;
const int N = 600 + 9;
const int M = 109;
const int MOD = int(1e9) + 7;
int n, m, k, need;
int st[N];
int d[N][N];
vector<int> g[N];
int mt[N];
bool u[N];
bool read() {
  if (!(cin >> n >> m >> k >> need)) return false;
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(n); ++j) d[i][j] = INF;
  for (int i = 0; i < int(n); ++i) d[i][i] = 0;
  for (int i = 0; i < int(k); ++i) cin >> st[i], --st[i];
  for (int i = 0; i < int(m); ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    --u, --v;
    d[u][v] = min(d[u][v], w);
    d[v][u] = min(d[v][u], w);
  }
  return true;
}
bool dfs(int v) {
  if (u[v]) return false;
  u[v] = true;
  for (int i = 0; i < int(int(g[v].size())); ++i) {
    int to = g[v][i];
    if (mt[to] == -1 || dfs(mt[to])) {
      mt[to] = v;
      return true;
    }
  }
  return false;
}
bool ok(int maxD) {
  for (int i = 0; i < int(n); ++i) g[i].clear();
  for (int i = 0; i < int(k); ++i)
    for (int j = 0; j < int(n); ++j)
      if (d[st[i]][j] <= maxD) g[i].push_back(j);
  memset(mt, -1, sizeof mt);
  int res = 0;
  for (int i = 0; i < int(k); ++i) {
    memset(u, 0, sizeof u);
    res += dfs(i);
  }
  return res >= need;
}
void solve() {
  for (int k = 0; k < int(n); ++k)
    for (int i = 0; i < int(n); ++i)
      for (int j = 0; j < int(n); ++j)
        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
  int l = -1, r = 1731311 + 100;
  while (r - l > 1) {
    int mid = (l + r) / 2;
    if (ok(mid))
      r = mid;
    else
      l = mid;
  }
  if (!ok(r))
    puts("-1");
  else
    cout << r << endl;
}
int main() {
  srand((unsigned int)time(NULL));
  cout << setprecision(15) << fixed;
  read();
  solve();
  return 0;
}

```