#include <bits/stdc++.h>
const int maxn = 1010;
const int maxm = 1e6 + 5;
const int INF = 0x3f3f3f3f;
using namespace std;
inline int read() {
  int x = 0, t = 1, c;
  while (!isdigit(c = getchar()))
    if (c == '-') t = -1;
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return x * t;
}
int V, E, N, K;
int head[maxn], cur[maxn], nx[maxm << 1], to[maxm << 1], flow[maxm << 1],
    cost[maxm << 1], ppp = 0;
int G[maxn][maxn], dis[maxn];
struct Dinic {
  int s, t;
  long long ans;
  void init() {
    memset(head, -1, sizeof(head));
    ppp = 0;
  }
  void AddEdge(int u, int v, int c) {
    to[ppp] = v;
    flow[ppp] = c;
    nx[ppp] = head[u];
    head[u] = ppp++;
    swap(u, v);
    to[ppp] = v;
    flow[ppp] = 0;
    nx[ppp] = head[u];
    head[u] = ppp++;
  }
  bool BFS() {
    memset(dis, -1, sizeof(dis));
    dis[s] = 1;
    queue<int> Q;
    Q.push(s);
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (int i = head[x]; ~i; i = nx[i]) {
        if (flow[i] && dis[to[i]] == -1) {
          dis[to[i]] = dis[x] + 1;
          Q.push(to[i]);
        }
      }
    }
    return dis[t] != -1;
  }
  int DFS(int x, int maxflow) {
    if (x == t || !maxflow) {
      ans += maxflow;
      return maxflow;
    }
    int ret = 0, f;
    for (int &i = cur[x]; ~i; i = nx[i]) {
      if (dis[to[i]] == dis[x] + 1 && (f = DFS(to[i], min(maxflow, flow[i])))) {
        ret += f;
        flow[i] -= f;
        flow[i ^ 1] += f;
        maxflow -= f;
        if (!maxflow) break;
      }
    }
    return ret;
  }
  long long solve(int source, int tank) {
    s = source;
    t = tank;
    ans = 0;
    while (BFS()) {
      memcpy(cur, head, sizeof(cur));
      DFS(s, INF);
    }
    return ans;
  }
} dinic;
void add_edge(int u, int v, int val) {
  nx[ppp] = head[u], cost[ppp] = val, to[ppp] = v, head[u] = ppp++;
  swap(u, v);
  nx[ppp] = head[u], cost[ppp] = val, to[ppp] = v, head[u] = ppp++;
}
void dijkstra(int s) {
  memset(dis, 0x3f, sizeof(dis));
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      q;
  dis[s] = 0;
  q.push(pair<int, int>(dis[s], s));
  while (!q.empty()) {
    pair<int, int> tmp = q.top();
    q.pop();
    int u = tmp.second;
    if (tmp.first > dis[u]) continue;
    for (int i = head[u]; ~i; i = nx[i]) {
      int v = to[i];
      int d = cost[i] + dis[u];
      if (dis[v] > d) {
        dis[v] = d;
        q.push(pair<int, int>(dis[v], v));
      }
    }
  }
}
int team[205];
void build(int mid, int s, int t) {
  dinic.init();
  for (int i = 1; i <= N; i++) {
    int tmp = team[i];
    dinic.AddEdge(s, i, 1);
    for (int j = 1; j <= V; j++) {
      if (G[tmp][j] <= mid) dinic.AddEdge(i, N + j, 1);
    }
  }
  for (int i = 1; i <= V; i++) dinic.AddEdge(N + i, t, 1);
}
int main() {
  V = read(), E = read(), N = read(), K = read();
  for (int i = 1; i <= N; i++) {
    team[i] = read();
  }
  dinic.init();
  for (int i = 0; i < E; i++) {
    int u = read(), v = read(), val = read();
    add_edge(u, v, val);
  }
  for (int i = 1; i <= V; i++) {
    dijkstra(i);
    for (int j = 1; j <= V; j++) {
      G[i][j] = dis[j];
    }
  }
  int front = 0, back = 1731313;
  int s = 0, t = N + V + 1;
  while (back > front) {
    int mid = (front + back) / 2;
    build(mid, s, t);
    int ans = dinic.solve(s, t);
    if (ans >= K)
      back = mid;
    else
      front = mid + 1;
  }
  if (front > 1731311)
    printf("-1\n");
  else
    printf("%d\n", front);
  return 0;
}

### Prompt

Please formulate 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>
const int maxn = 1010;
const int maxm = 1e6 + 5;
const int INF = 0x3f3f3f3f;
using namespace std;
inline int read() {
  int x = 0, t = 1, c;
  while (!isdigit(c = getchar()))
    if (c == '-') t = -1;
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return x * t;
}
int V, E, N, K;
int head[maxn], cur[maxn], nx[maxm << 1], to[maxm << 1], flow[maxm << 1],
    cost[maxm << 1], ppp = 0;
int G[maxn][maxn], dis[maxn];
struct Dinic {
  int s, t;
  long long ans;
  void init() {
    memset(head, -1, sizeof(head));
    ppp = 0;
  }
  void AddEdge(int u, int v, int c) {
    to[ppp] = v;
    flow[ppp] = c;
    nx[ppp] = head[u];
    head[u] = ppp++;
    swap(u, v);
    to[ppp] = v;
    flow[ppp] = 0;
    nx[ppp] = head[u];
    head[u] = ppp++;
  }
  bool BFS() {
    memset(dis, -1, sizeof(dis));
    dis[s] = 1;
    queue<int> Q;
    Q.push(s);
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (int i = head[x]; ~i; i = nx[i]) {
        if (flow[i] && dis[to[i]] == -1) {
          dis[to[i]] = dis[x] + 1;
          Q.push(to[i]);
        }
      }
    }
    return dis[t] != -1;
  }
  int DFS(int x, int maxflow) {
    if (x == t || !maxflow) {
      ans += maxflow;
      return maxflow;
    }
    int ret = 0, f;
    for (int &i = cur[x]; ~i; i = nx[i]) {
      if (dis[to[i]] == dis[x] + 1 && (f = DFS(to[i], min(maxflow, flow[i])))) {
        ret += f;
        flow[i] -= f;
        flow[i ^ 1] += f;
        maxflow -= f;
        if (!maxflow) break;
      }
    }
    return ret;
  }
  long long solve(int source, int tank) {
    s = source;
    t = tank;
    ans = 0;
    while (BFS()) {
      memcpy(cur, head, sizeof(cur));
      DFS(s, INF);
    }
    return ans;
  }
} dinic;
void add_edge(int u, int v, int val) {
  nx[ppp] = head[u], cost[ppp] = val, to[ppp] = v, head[u] = ppp++;
  swap(u, v);
  nx[ppp] = head[u], cost[ppp] = val, to[ppp] = v, head[u] = ppp++;
}
void dijkstra(int s) {
  memset(dis, 0x3f, sizeof(dis));
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      q;
  dis[s] = 0;
  q.push(pair<int, int>(dis[s], s));
  while (!q.empty()) {
    pair<int, int> tmp = q.top();
    q.pop();
    int u = tmp.second;
    if (tmp.first > dis[u]) continue;
    for (int i = head[u]; ~i; i = nx[i]) {
      int v = to[i];
      int d = cost[i] + dis[u];
      if (dis[v] > d) {
        dis[v] = d;
        q.push(pair<int, int>(dis[v], v));
      }
    }
  }
}
int team[205];
void build(int mid, int s, int t) {
  dinic.init();
  for (int i = 1; i <= N; i++) {
    int tmp = team[i];
    dinic.AddEdge(s, i, 1);
    for (int j = 1; j <= V; j++) {
      if (G[tmp][j] <= mid) dinic.AddEdge(i, N + j, 1);
    }
  }
  for (int i = 1; i <= V; i++) dinic.AddEdge(N + i, t, 1);
}
int main() {
  V = read(), E = read(), N = read(), K = read();
  for (int i = 1; i <= N; i++) {
    team[i] = read();
  }
  dinic.init();
  for (int i = 0; i < E; i++) {
    int u = read(), v = read(), val = read();
    add_edge(u, v, val);
  }
  for (int i = 1; i <= V; i++) {
    dijkstra(i);
    for (int j = 1; j <= V; j++) {
      G[i][j] = dis[j];
    }
  }
  int front = 0, back = 1731313;
  int s = 0, t = N + V + 1;
  while (back > front) {
    int mid = (front + back) / 2;
    build(mid, s, t);
    int ans = dinic.solve(s, t);
    if (ans >= K)
      back = mid;
    else
      front = mid + 1;
  }
  if (front > 1731311)
    printf("-1\n");
  else
    printf("%d\n", front);
  return 0;
}

```