#include <bits/stdc++.h>
using namespace std;
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
struct edge {
  int a, b, cap, flow;
  edge() {}
  edge(int a, int b, int cap, int flow) {
    this->a = a;
    this->b = b;
    this->cap = cap;
    this->flow = flow;
  }
};
int src, sink, ptr[1205], dist[1205];
vector<edge> edge_list;
vector<int> adj[1205];
void add_edge(int a, int b, int cap) {
  adj[a].push_back((int)edge_list.size());
  edge_list.push_back(edge(a, b, cap, 0));
  adj[b].push_back((int)edge_list.size());
  edge_list.push_back(edge(b, a, 0, 0));
}
bool bfs() {
  memset(dist, -1, sizeof(dist));
  queue<int> qq;
  dist[src] = 0;
  qq.push(src);
  while (!qq.empty()) {
    int v = qq.front();
    qq.pop();
    for (int i = 0; i < (int)adj[v].size(); i++) {
      int e = adj[v][i];
      int vv = edge_list[e].b;
      if (dist[vv] == -1 && edge_list[e].flow < edge_list[e].cap) {
        dist[vv] = dist[v] + 1;
        qq.push(vv);
      }
    }
  }
  return dist[sink] != -1;
}
int dfs(int v, int flow) {
  if (v == sink || flow == 0) return flow;
  for (; ptr[v] < (int)adj[v].size(); ptr[v]++) {
    int id = adj[v][ptr[v]];
    int vv = edge_list[id].b;
    if (dist[vv] != dist[v] + 1) continue;
    int pushed = dfs(vv, min(flow, edge_list[id].cap - edge_list[id].flow));
    if (pushed) {
      edge_list[id].flow += pushed;
      edge_list[1 ^ id].flow -= pushed;
      return pushed;
    }
  }
  return 0;
}
int dinitz() {
  int flow = 0;
  while (1) {
    if (!bfs()) break;
    memset(ptr, 0, sizeof(ptr));
    while (int pushed = dfs(src, (1000000007LL))) flow += pushed;
  }
  return flow;
}
int n, e, cnt[605], adj_[605][605];
int count_reachable(int mid) {
  src = 0;
  sink = 2 * n + 1;
  edge_list.clear();
  for (int i = src; i <= sink; i++) {
    adj[i].clear();
  }
  for (int i = 1; i <= n; i++) {
    add_edge(src, i, cnt[i]);
    add_edge(n + i, sink, 1);
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (adj_[i][j] <= mid) add_edge(i, n + j, 1);
  int ans = dinitz();
  return ans;
}
int main() {
  int sz, k;
  scanf("%d%d%d%d", &n, &e, &sz, &k);
  while (sz--) {
    int a;
    scanf("%d", &a);
    cnt[a]++;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      adj_[i][j] = (1000000007LL);
      adj_[i][i] = 0;
    }
  while (e--) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    adj_[a][b] = adj_[b][a] = min(adj_[a][b], c);
  }
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        adj_[i][j] = min(adj_[i][j], adj_[i][k] + adj_[k][j]);
  int lo = 0, hi = 1731312, ans = (1000000007LL);
  while (lo <= hi) {
    int mid = (lo + hi) / 2;
    if (count_reachable(mid) >= k) {
      ans = min(ans, mid);
      hi = mid - 1;
    } else {
      lo = mid + 1;
    }
  }
  if (ans == (1000000007LL)) ans = -1;
  printf("%d\n", ans);
  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;
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
struct edge {
  int a, b, cap, flow;
  edge() {}
  edge(int a, int b, int cap, int flow) {
    this->a = a;
    this->b = b;
    this->cap = cap;
    this->flow = flow;
  }
};
int src, sink, ptr[1205], dist[1205];
vector<edge> edge_list;
vector<int> adj[1205];
void add_edge(int a, int b, int cap) {
  adj[a].push_back((int)edge_list.size());
  edge_list.push_back(edge(a, b, cap, 0));
  adj[b].push_back((int)edge_list.size());
  edge_list.push_back(edge(b, a, 0, 0));
}
bool bfs() {
  memset(dist, -1, sizeof(dist));
  queue<int> qq;
  dist[src] = 0;
  qq.push(src);
  while (!qq.empty()) {
    int v = qq.front();
    qq.pop();
    for (int i = 0; i < (int)adj[v].size(); i++) {
      int e = adj[v][i];
      int vv = edge_list[e].b;
      if (dist[vv] == -1 && edge_list[e].flow < edge_list[e].cap) {
        dist[vv] = dist[v] + 1;
        qq.push(vv);
      }
    }
  }
  return dist[sink] != -1;
}
int dfs(int v, int flow) {
  if (v == sink || flow == 0) return flow;
  for (; ptr[v] < (int)adj[v].size(); ptr[v]++) {
    int id = adj[v][ptr[v]];
    int vv = edge_list[id].b;
    if (dist[vv] != dist[v] + 1) continue;
    int pushed = dfs(vv, min(flow, edge_list[id].cap - edge_list[id].flow));
    if (pushed) {
      edge_list[id].flow += pushed;
      edge_list[1 ^ id].flow -= pushed;
      return pushed;
    }
  }
  return 0;
}
int dinitz() {
  int flow = 0;
  while (1) {
    if (!bfs()) break;
    memset(ptr, 0, sizeof(ptr));
    while (int pushed = dfs(src, (1000000007LL))) flow += pushed;
  }
  return flow;
}
int n, e, cnt[605], adj_[605][605];
int count_reachable(int mid) {
  src = 0;
  sink = 2 * n + 1;
  edge_list.clear();
  for (int i = src; i <= sink; i++) {
    adj[i].clear();
  }
  for (int i = 1; i <= n; i++) {
    add_edge(src, i, cnt[i]);
    add_edge(n + i, sink, 1);
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (adj_[i][j] <= mid) add_edge(i, n + j, 1);
  int ans = dinitz();
  return ans;
}
int main() {
  int sz, k;
  scanf("%d%d%d%d", &n, &e, &sz, &k);
  while (sz--) {
    int a;
    scanf("%d", &a);
    cnt[a]++;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      adj_[i][j] = (1000000007LL);
      adj_[i][i] = 0;
    }
  while (e--) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    adj_[a][b] = adj_[b][a] = min(adj_[a][b], c);
  }
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        adj_[i][j] = min(adj_[i][j], adj_[i][k] + adj_[k][j]);
  int lo = 0, hi = 1731312, ans = (1000000007LL);
  while (lo <= hi) {
    int mid = (lo + hi) / 2;
    if (count_reachable(mid) >= k) {
      ans = min(ans, mid);
      hi = mid - 1;
    } else {
      lo = mid + 1;
    }
  }
  if (ans == (1000000007LL)) ans = -1;
  printf("%d\n", ans);
  return 0;
}

```