#include <bits/stdc++.h>
using namespace std;
const int maxn = 10005;
const int maxm = 1000005;
const int INF = 0x3f3f3f3f;
int head[maxn], cur[maxn], nx[maxm << 1], to[maxm << 1], flow[maxm << 1],
    ppp = 0;
struct Dinic {
  int dis[maxn];
  int s, t;
  int 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[t] = 1;
    queue<int> Q;
    Q.push(t);
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (int i = head[x]; ~i; i = nx[i]) {
        if (flow[i ^ 1] && dis[to[i]] == -1) {
          dis[to[i]] = dis[x] + 1;
          Q.push(to[i]);
        }
      }
    }
    return dis[s] != -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;
  }
  int solve(int source, int tank) {
    s = source;
    t = tank;
    ans = 0;
    while (BFS()) {
      memcpy(cur, head, sizeof(cur));
      DFS(s, INF);
    }
    return ans;
  }
} dinic;
vector<pair<int, int> > edge[605];
int cnt[605], dis[605][605];
int n, m, e;
void floyd() {
  for (int k = 1; k <= m; k++) {
    for (int i = 1; i <= m; i++) {
      for (int j = 1; j <= m; j++) {
        dis[i][j] = min(dis[i][k] + dis[k][j], dis[i][j]);
      }
    }
  }
}
int main() {
  int k;
  ios::sync_with_stdio(0);
  cin >> m >> e >> n >> k;
  for (int i = 1, tmp; i <= n; i++) {
    cin >> tmp;
    cnt[tmp]++;
  }
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= m; j++) {
      dis[i][j] = INF;
    }
  }
  for (int i = 0, u, v, val; i < e; i++) {
    cin >> u >> v >> val;
    if (u == v) continue;
    if (dis[u][v] > val) dis[u][v] = dis[v][u] = val;
  }
  floyd();
  int front = 0, back = 1731311;
  while (front <= back) {
    int mid = (front + back) / 2;
    dinic.init();
    int s = 0, e = 2 * m + 1;
    for (int i = 1; i <= m; i++) {
      dinic.AddEdge(s, i, cnt[i]);
      dinic.AddEdge(m + i, e, 1);
    }
    for (int i = 1; i <= m; i++) {
      for (int j = 1; j <= m; j++) {
        if (dis[i][j] <= mid || i == j) dinic.AddEdge(i, m + j, INF);
      }
    }
    int Max_Flow = dinic.solve(s, e);
    if (Max_Flow >= k) {
      back = mid - 1;
    } else {
      front = mid + 1;
    }
  }
  if (front <= 1731311)
    cout << front << endl;
  else
    cout << -1 << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
const int maxn = 10005;
const int maxm = 1000005;
const int INF = 0x3f3f3f3f;
int head[maxn], cur[maxn], nx[maxm << 1], to[maxm << 1], flow[maxm << 1],
    ppp = 0;
struct Dinic {
  int dis[maxn];
  int s, t;
  int 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[t] = 1;
    queue<int> Q;
    Q.push(t);
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (int i = head[x]; ~i; i = nx[i]) {
        if (flow[i ^ 1] && dis[to[i]] == -1) {
          dis[to[i]] = dis[x] + 1;
          Q.push(to[i]);
        }
      }
    }
    return dis[s] != -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;
  }
  int solve(int source, int tank) {
    s = source;
    t = tank;
    ans = 0;
    while (BFS()) {
      memcpy(cur, head, sizeof(cur));
      DFS(s, INF);
    }
    return ans;
  }
} dinic;
vector<pair<int, int> > edge[605];
int cnt[605], dis[605][605];
int n, m, e;
void floyd() {
  for (int k = 1; k <= m; k++) {
    for (int i = 1; i <= m; i++) {
      for (int j = 1; j <= m; j++) {
        dis[i][j] = min(dis[i][k] + dis[k][j], dis[i][j]);
      }
    }
  }
}
int main() {
  int k;
  ios::sync_with_stdio(0);
  cin >> m >> e >> n >> k;
  for (int i = 1, tmp; i <= n; i++) {
    cin >> tmp;
    cnt[tmp]++;
  }
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= m; j++) {
      dis[i][j] = INF;
    }
  }
  for (int i = 0, u, v, val; i < e; i++) {
    cin >> u >> v >> val;
    if (u == v) continue;
    if (dis[u][v] > val) dis[u][v] = dis[v][u] = val;
  }
  floyd();
  int front = 0, back = 1731311;
  while (front <= back) {
    int mid = (front + back) / 2;
    dinic.init();
    int s = 0, e = 2 * m + 1;
    for (int i = 1; i <= m; i++) {
      dinic.AddEdge(s, i, cnt[i]);
      dinic.AddEdge(m + i, e, 1);
    }
    for (int i = 1; i <= m; i++) {
      for (int j = 1; j <= m; j++) {
        if (dis[i][j] <= mid || i == j) dinic.AddEdge(i, m + j, INF);
      }
    }
    int Max_Flow = dinic.solve(s, e);
    if (Max_Flow >= k) {
      back = mid - 1;
    } else {
      front = mid + 1;
    }
  }
  if (front <= 1731311)
    cout << front << endl;
  else
    cout << -1 << endl;
  return 0;
}

```