#include <bits/stdc++.h>
using namespace std;
struct edge {
  int v;
  edge() { v = 0; }
  edge(int v) { this->v = v; }
};
int n, m, N, K, S, T;
vector<edge> G[1205];
vector<int> f[1206];
vector<pair<int, int> > adj[605];
int indeg[607];
int dist[1205], p[1205];
int dp[1205][1205];
int idx;
int my[1205], mx[1205], vs[1205];
void precalc() {
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) dp[i][j] = 1e9;
  for (int i = 1; i <= n; i++) {
    dp[i][i] = 0;
    priority_queue<pair<int, int> > q;
    q.push(make_pair(0, i));
    while (!q.empty()) {
      int u = q.top().second;
      int z = -q.top().first;
      q.pop();
      if (dp[i][u] != z) continue;
      for (int j = adj[u].size() - 1; j >= 0; j--) {
        int v = adj[u][j].first;
        int cost = adj[u][j].second;
        if (dp[i][v] > dp[i][u] + cost) {
          dp[i][v] = dp[i][u] + cost;
          q.push(make_pair(-dp[i][v], v));
        }
      }
    }
  }
}
int aug(int u) {
  if (vs[u]) return 0;
  vs[u] = 1;
  for (int i = f[u].size() - 1; i >= 0; i--) {
    int v = f[u][i];
    if (my[v] == -1 || aug(my[v])) {
      mx[u] = v;
      my[v] = u;
      return 1;
    }
  }
  return 0;
}
int solve(int lim) {
  for (int i = 1; i <= N; i++) {
    f[i].clear();
    mx[i] = -1;
  }
  for (int i = 1; i <= n; i++) my[i] = -1;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= n; j++) {
      if (dp[indeg[i]][j] <= lim) f[i].push_back(j);
    }
  }
  int flow = 0;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) vs[j] = 0;
    if (mx[i] == -1) flow += aug(i);
  }
  return flow;
}
int bs(int lo, int hi) {
  while (1) {
    if (hi - lo <= 1) {
      if (solve(lo) >= K) return lo;
      if (solve(hi) >= K) return hi;
      return -1;
    }
    int mid = (lo + hi) / 2;
    if (solve(mid) >= K)
      hi = mid;
    else
      lo = mid;
  }
}
int main() {
  scanf("%d%d%d%d", &n, &m, &N, &K);
  S = 0, T = n + N + 1;
  for (int i = 1; i <= N; i++) {
    int x;
    scanf("%d", &x);
    indeg[i] = x;
  }
  for (int i = 0; i < m; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    adj[x].push_back(make_pair(y, z));
    adj[y].push_back(make_pair(x, z));
  }
  precalc();
  printf("%d\n", bs(0, 1731312));
  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;
struct edge {
  int v;
  edge() { v = 0; }
  edge(int v) { this->v = v; }
};
int n, m, N, K, S, T;
vector<edge> G[1205];
vector<int> f[1206];
vector<pair<int, int> > adj[605];
int indeg[607];
int dist[1205], p[1205];
int dp[1205][1205];
int idx;
int my[1205], mx[1205], vs[1205];
void precalc() {
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) dp[i][j] = 1e9;
  for (int i = 1; i <= n; i++) {
    dp[i][i] = 0;
    priority_queue<pair<int, int> > q;
    q.push(make_pair(0, i));
    while (!q.empty()) {
      int u = q.top().second;
      int z = -q.top().first;
      q.pop();
      if (dp[i][u] != z) continue;
      for (int j = adj[u].size() - 1; j >= 0; j--) {
        int v = adj[u][j].first;
        int cost = adj[u][j].second;
        if (dp[i][v] > dp[i][u] + cost) {
          dp[i][v] = dp[i][u] + cost;
          q.push(make_pair(-dp[i][v], v));
        }
      }
    }
  }
}
int aug(int u) {
  if (vs[u]) return 0;
  vs[u] = 1;
  for (int i = f[u].size() - 1; i >= 0; i--) {
    int v = f[u][i];
    if (my[v] == -1 || aug(my[v])) {
      mx[u] = v;
      my[v] = u;
      return 1;
    }
  }
  return 0;
}
int solve(int lim) {
  for (int i = 1; i <= N; i++) {
    f[i].clear();
    mx[i] = -1;
  }
  for (int i = 1; i <= n; i++) my[i] = -1;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= n; j++) {
      if (dp[indeg[i]][j] <= lim) f[i].push_back(j);
    }
  }
  int flow = 0;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) vs[j] = 0;
    if (mx[i] == -1) flow += aug(i);
  }
  return flow;
}
int bs(int lo, int hi) {
  while (1) {
    if (hi - lo <= 1) {
      if (solve(lo) >= K) return lo;
      if (solve(hi) >= K) return hi;
      return -1;
    }
    int mid = (lo + hi) / 2;
    if (solve(mid) >= K)
      hi = mid;
    else
      lo = mid;
  }
}
int main() {
  scanf("%d%d%d%d", &n, &m, &N, &K);
  S = 0, T = n + N + 1;
  for (int i = 1; i <= N; i++) {
    int x;
    scanf("%d", &x);
    indeg[i] = x;
  }
  for (int i = 0; i < m; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    adj[x].push_back(make_pair(y, z));
    adj[y].push_back(make_pair(x, z));
  }
  precalc();
  printf("%d\n", bs(0, 1731312));
  return 0;
}

```