#include <bits/stdc++.h>
using namespace std;
struct MaxFlow {
  int T;
  struct node {
    int to, nxt, cap;
  } ed[811 * 811 << 1];
  int head[811], cur[811], cnt, ans;
  int h[811];
  void addedge(int u, int v, int cap) {
    ed[cnt].to = v, ed[cnt].cap = cap, ed[cnt].nxt = head[u], head[u] = cnt++;
    ed[cnt].to = u, ed[cnt].cap = 0, ed[cnt].nxt = head[v], head[v] = cnt++;
  }
  bool bfs() {
    memset(h, -1, sizeof(h));
    queue<int> q;
    h[1] = 0, q.push(1);
    while (!q.empty()) {
      int now = q.front();
      q.pop();
      for (int i = head[now]; ~i; i = ed[i].nxt) {
        if (ed[i].cap && h[ed[i].to] == -1) {
          h[ed[i].to] = h[now] + 1;
          q.push(ed[i].to);
        }
      }
    }
    return h[T] != -1;
  }
  int dfs(int x, int f) {
    if (x == T || f == 0) return f;
    int w, used = 0;
    for (int i = cur[x]; ~i; i = ed[i].nxt) {
      if (h[ed[i].to] == h[x] + 1) {
        w = f - used;
        w = dfs(ed[i].to, min(w, ed[i].cap));
        ed[i].cap -= w;
        ed[i ^ 1].cap += w;
        if (ed[i].cap) cur[x] = i;
        used += w;
        if (used == f) return f;
      }
    }
    if (!used) h[x] = -1;
    return used;
  }
  void dinic() {
    while (bfs()) {
      for (int i = 1; i <= T; i++) {
        cur[i] = head[i];
      }
      ans += dfs(1, 100000005);
    }
  }
  void init(int t) {
    memset(head, -1, sizeof(head));
    ans = cnt = 0;
    T = t;
  }
} flow;
int dis[811][811], st[811];
int main() {
  int v, e, n, k;
  scanf("%d%d%d%d", &v, &e, &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &st[i]);
  for (int i = 1; i <= v; i++)
    for (int j = i + 1; j <= v; j++) dis[i][j] = dis[j][i] = 100000005;
  for (int i = 1; i <= e; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    dis[a][b] = min(dis[a][b], c);
    dis[b][a] = dis[a][b];
  }
  for (int x = 1; x <= v; x++) {
    for (int i = 1; i <= v; i++) {
      for (int j = 1; j <= v; j++) {
        dis[i][j] = min(dis[i][x] + dis[x][j], dis[i][j]);
      }
    }
  }
  int l = 0, r = 1731311, ct = -1;
  while (l <= r) {
    int mid = l + r >> 1;
    flow.init(1 + n + v + 2);
    for (int i = 1; i <= n; i++) {
      flow.addedge(1, i + 1, 1);
      for (int j = 1; j <= v; j++) {
        if (dis[st[i]][j] <= mid) {
          flow.addedge(i + 1, 1 + j + n, 1);
        }
      }
    }
    for (int i = 1; i <= v; i++) flow.addedge(1 + i + n, 1 + v + n + 1, 1);
    flow.addedge(1 + v + n + 1, 1 + v + n + 2, k);
    flow.dinic();
    if (flow.ans == k) {
      ct = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  printf("%d\n", ct);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 MaxFlow {
  int T;
  struct node {
    int to, nxt, cap;
  } ed[811 * 811 << 1];
  int head[811], cur[811], cnt, ans;
  int h[811];
  void addedge(int u, int v, int cap) {
    ed[cnt].to = v, ed[cnt].cap = cap, ed[cnt].nxt = head[u], head[u] = cnt++;
    ed[cnt].to = u, ed[cnt].cap = 0, ed[cnt].nxt = head[v], head[v] = cnt++;
  }
  bool bfs() {
    memset(h, -1, sizeof(h));
    queue<int> q;
    h[1] = 0, q.push(1);
    while (!q.empty()) {
      int now = q.front();
      q.pop();
      for (int i = head[now]; ~i; i = ed[i].nxt) {
        if (ed[i].cap && h[ed[i].to] == -1) {
          h[ed[i].to] = h[now] + 1;
          q.push(ed[i].to);
        }
      }
    }
    return h[T] != -1;
  }
  int dfs(int x, int f) {
    if (x == T || f == 0) return f;
    int w, used = 0;
    for (int i = cur[x]; ~i; i = ed[i].nxt) {
      if (h[ed[i].to] == h[x] + 1) {
        w = f - used;
        w = dfs(ed[i].to, min(w, ed[i].cap));
        ed[i].cap -= w;
        ed[i ^ 1].cap += w;
        if (ed[i].cap) cur[x] = i;
        used += w;
        if (used == f) return f;
      }
    }
    if (!used) h[x] = -1;
    return used;
  }
  void dinic() {
    while (bfs()) {
      for (int i = 1; i <= T; i++) {
        cur[i] = head[i];
      }
      ans += dfs(1, 100000005);
    }
  }
  void init(int t) {
    memset(head, -1, sizeof(head));
    ans = cnt = 0;
    T = t;
  }
} flow;
int dis[811][811], st[811];
int main() {
  int v, e, n, k;
  scanf("%d%d%d%d", &v, &e, &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &st[i]);
  for (int i = 1; i <= v; i++)
    for (int j = i + 1; j <= v; j++) dis[i][j] = dis[j][i] = 100000005;
  for (int i = 1; i <= e; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    dis[a][b] = min(dis[a][b], c);
    dis[b][a] = dis[a][b];
  }
  for (int x = 1; x <= v; x++) {
    for (int i = 1; i <= v; i++) {
      for (int j = 1; j <= v; j++) {
        dis[i][j] = min(dis[i][x] + dis[x][j], dis[i][j]);
      }
    }
  }
  int l = 0, r = 1731311, ct = -1;
  while (l <= r) {
    int mid = l + r >> 1;
    flow.init(1 + n + v + 2);
    for (int i = 1; i <= n; i++) {
      flow.addedge(1, i + 1, 1);
      for (int j = 1; j <= v; j++) {
        if (dis[st[i]][j] <= mid) {
          flow.addedge(i + 1, 1 + j + n, 1);
        }
      }
    }
    for (int i = 1; i <= v; i++) flow.addedge(1 + i + n, 1 + v + n + 1, 1);
    flow.addedge(1 + v + n + 1, 1 + v + n + 2, k);
    flow.dinic();
    if (flow.ans == k) {
      ct = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  printf("%d\n", ct);
  return 0;
}

```