#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
const int inf = 1000 * 1000 * 1000;
const int mod = 1000 * 1000 * 1000 + 7;
struct edge {
  int to, cap, flow;
};
int S, T;
vector<edge> G;
vector<int> g[N];
int qan[N], a[N], pos[N];
string s;
int n, m, x, k;
int mas[N], id[N], ind[N], parent[N], d[N];
int dist[N][N];
void clean() {
  G.clear();
  for (int i = 0; i <= T; i++) g[i].clear();
}
void add_edge(int a, int b, int c) {
  edge e1 = {b, c, 0};
  edge e2 = {a, 0, 0};
  g[a].push_back((int)G.size());
  G.push_back(e1);
  g[b].push_back((int)G.size());
  G.push_back(e2);
}
bool bfs() {
  queue<int> q;
  for (int i = 0; i <= T; i++) d[i] = -1;
  q.push(S);
  d[S] = 0;
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < (int)g[v].size(); i++) {
      edge e = G[g[v][i]];
      if (e.cap == e.flow) continue;
      int to = e.to;
      if (d[to] == -1) {
        q.push(to);
        d[to] = d[v] + 1;
        parent[to] = v;
        ind[to] = i;
      }
    }
    if (d[T] != -1) break;
  }
  return (d[T] != -1);
}
int dfs(int v, int flow) {
  if (flow == 0) return 0;
  if (v == T) return flow;
  for (; pos[v] < (int)g[v].size(); pos[v]++) {
    edge e = G[g[v][pos[v]]];
    if (e.cap == e.flow) continue;
    int to = e.to;
    if (d[to] != d[v] + 1) continue;
    int push = dfs(to, min(flow, e.cap - e.flow));
    if (push) {
      G[g[v][pos[v]]].flow += push;
      G[g[v][pos[v]] ^ 1].flow -= push;
      return push;
    }
  }
  return 0;
}
bool dinic() {
  int flow = 0;
  while (bfs()) {
    for (int i = 0; i <= T; i++) pos[i] = 0;
    while (int push = dfs(S, inf)) {
      flow += push;
    }
  }
  return (flow >= x);
}
int main() {
  cin >> n >> m >> k >> x;
  for (int i = 1; i <= k; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      dist[i][j] = inf;
    }
  }
  for (int i = 1; i <= m; i++) {
    int aa, bb, tt;
    scanf("%d%d%d", &aa, &bb, &tt);
    dist[aa][bb] = min(dist[aa][bb], tt);
    dist[bb][aa] = min(dist[bb][aa], tt);
  }
  for (int k = 1; k <= n; k++) {
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
      }
    }
  }
  S = 0;
  T = k + n + 1;
  int ina = 0, inb = 1731311, answ = -1;
  while (ina <= inb) {
    int mid = (ina + inb) / 2;
    clean();
    for (int i = 1; i <= k; i++) {
      add_edge(S, i, 1);
      for (int j = 1; j <= n; j++) {
        if (dist[a[i]][j] <= mid) add_edge(i, j + k, 1);
      }
    }
    for (int i = k + 1; i <= k + n; i++) add_edge(i, T, 1);
    if (dinic()) {
      answ = mid;
      inb = mid - 1;
    } else
      ina = mid + 1;
  }
  cout << answ << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int N = 1005;
const int inf = 1000 * 1000 * 1000;
const int mod = 1000 * 1000 * 1000 + 7;
struct edge {
  int to, cap, flow;
};
int S, T;
vector<edge> G;
vector<int> g[N];
int qan[N], a[N], pos[N];
string s;
int n, m, x, k;
int mas[N], id[N], ind[N], parent[N], d[N];
int dist[N][N];
void clean() {
  G.clear();
  for (int i = 0; i <= T; i++) g[i].clear();
}
void add_edge(int a, int b, int c) {
  edge e1 = {b, c, 0};
  edge e2 = {a, 0, 0};
  g[a].push_back((int)G.size());
  G.push_back(e1);
  g[b].push_back((int)G.size());
  G.push_back(e2);
}
bool bfs() {
  queue<int> q;
  for (int i = 0; i <= T; i++) d[i] = -1;
  q.push(S);
  d[S] = 0;
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < (int)g[v].size(); i++) {
      edge e = G[g[v][i]];
      if (e.cap == e.flow) continue;
      int to = e.to;
      if (d[to] == -1) {
        q.push(to);
        d[to] = d[v] + 1;
        parent[to] = v;
        ind[to] = i;
      }
    }
    if (d[T] != -1) break;
  }
  return (d[T] != -1);
}
int dfs(int v, int flow) {
  if (flow == 0) return 0;
  if (v == T) return flow;
  for (; pos[v] < (int)g[v].size(); pos[v]++) {
    edge e = G[g[v][pos[v]]];
    if (e.cap == e.flow) continue;
    int to = e.to;
    if (d[to] != d[v] + 1) continue;
    int push = dfs(to, min(flow, e.cap - e.flow));
    if (push) {
      G[g[v][pos[v]]].flow += push;
      G[g[v][pos[v]] ^ 1].flow -= push;
      return push;
    }
  }
  return 0;
}
bool dinic() {
  int flow = 0;
  while (bfs()) {
    for (int i = 0; i <= T; i++) pos[i] = 0;
    while (int push = dfs(S, inf)) {
      flow += push;
    }
  }
  return (flow >= x);
}
int main() {
  cin >> n >> m >> k >> x;
  for (int i = 1; i <= k; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      dist[i][j] = inf;
    }
  }
  for (int i = 1; i <= m; i++) {
    int aa, bb, tt;
    scanf("%d%d%d", &aa, &bb, &tt);
    dist[aa][bb] = min(dist[aa][bb], tt);
    dist[bb][aa] = min(dist[bb][aa], tt);
  }
  for (int k = 1; k <= n; k++) {
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
      }
    }
  }
  S = 0;
  T = k + n + 1;
  int ina = 0, inb = 1731311, answ = -1;
  while (ina <= inb) {
    int mid = (ina + inb) / 2;
    clean();
    for (int i = 1; i <= k; i++) {
      add_edge(S, i, 1);
      for (int j = 1; j <= n; j++) {
        if (dist[a[i]][j] <= mid) add_edge(i, j + k, 1);
      }
    }
    for (int i = k + 1; i <= k + n; i++) add_edge(i, T, 1);
    if (dinic()) {
      answ = mid;
      inb = mid - 1;
    } else
      ina = mid + 1;
  }
  cout << answ << endl;
  return 0;
}

```