#include <bits/stdc++.h>
using namespace std;
struct edge {
  int v, nxt, c;
} e[1500005];
int p[1305], eid, pp[1305];
void init() {
  memset(p, -1, sizeof p);
  eid = 0;
}
void insert(int u, int v, int c) {
  e[eid].v = v;
  e[eid].c = c;
  e[eid].nxt = p[u];
  p[u] = eid++;
}
void add(int u, int v, int c) {
  insert(u, v, c);
  insert(v, u, 0);
}
int d[1305], S, T, n, m, sum[1305], h[1305], SS, TT, r[1305], g[1305][1305], k,
    lim, a1, a2, an, b1, b2, bn, a[505][505], ss[1305], sp[1305], nn;
queue<int> q;
int bfs() {
  for (int i = 1; i <= n * 2 + 3; i++) pp[i] = p[i], d[i] = -1;
  d[S] = 0;
  q.push(S);
  while (q.size()) {
    int u = q.front();
    q.pop();
    for (int i = p[u]; i + 1; i = e[i].nxt) {
      int v = e[i].v, c = e[i].c;
      if (d[v] == -1 && c) {
        d[v] = d[u] + 1;
        q.push(v);
      }
    }
  }
  return d[T] != -1;
}
int dfs(int u, int flow) {
  if (u == T) return flow;
  int ret = 0;
  for (int &i = pp[u]; i + 1; i = e[i].nxt) {
    int v = e[i].v, c = e[i].c;
    if (c && d[v] == d[u] + 1) {
      int tmp = dfs(v, min(flow, c));
      e[i].c -= tmp, e[i ^ 1].c += tmp;
      flow -= tmp, ret += tmp;
      if (!flow) break;
    }
  }
  if (!ret) d[u] = -1;
  return ret;
}
int Dinic() {
  int ret = 0;
  for (; bfs();) ret += dfs(S, 23333333);
  return ret;
}
int check(int mid) {
  init();
  S = n * 2 + 1, T = n * 2 + 2;
  for (int i = 1; i <= n; i++) add(S, i, sum[i]), add(i + n, T, 1);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (g[i][j] <= mid) add(i, j + n, 23333333);
  return Dinic() >= k;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &nn, &k);
  for (int i = 1; i <= nn; i++) {
    int x;
    scanf("%d", &x);
    sum[x]++;
  }
  memset(g, 0x3f, sizeof g);
  for (int i = 1; i <= m; i++) {
    int u, v, c;
    scanf("%d%d%d", &u, &v, &c);
    g[v][u] = g[u][v] = min(g[u][v], c);
  }
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        g[i][j] = min(g[i][j], g[i][k] + g[k][j]), g[i][i] = 0;
  int l = -1, r = 2e6;
  while (l + 1 < r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  printf("%d", r == 2e6 ? -1 : r);
  return 0;
}

### Prompt

Your task is to create 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;
struct edge {
  int v, nxt, c;
} e[1500005];
int p[1305], eid, pp[1305];
void init() {
  memset(p, -1, sizeof p);
  eid = 0;
}
void insert(int u, int v, int c) {
  e[eid].v = v;
  e[eid].c = c;
  e[eid].nxt = p[u];
  p[u] = eid++;
}
void add(int u, int v, int c) {
  insert(u, v, c);
  insert(v, u, 0);
}
int d[1305], S, T, n, m, sum[1305], h[1305], SS, TT, r[1305], g[1305][1305], k,
    lim, a1, a2, an, b1, b2, bn, a[505][505], ss[1305], sp[1305], nn;
queue<int> q;
int bfs() {
  for (int i = 1; i <= n * 2 + 3; i++) pp[i] = p[i], d[i] = -1;
  d[S] = 0;
  q.push(S);
  while (q.size()) {
    int u = q.front();
    q.pop();
    for (int i = p[u]; i + 1; i = e[i].nxt) {
      int v = e[i].v, c = e[i].c;
      if (d[v] == -1 && c) {
        d[v] = d[u] + 1;
        q.push(v);
      }
    }
  }
  return d[T] != -1;
}
int dfs(int u, int flow) {
  if (u == T) return flow;
  int ret = 0;
  for (int &i = pp[u]; i + 1; i = e[i].nxt) {
    int v = e[i].v, c = e[i].c;
    if (c && d[v] == d[u] + 1) {
      int tmp = dfs(v, min(flow, c));
      e[i].c -= tmp, e[i ^ 1].c += tmp;
      flow -= tmp, ret += tmp;
      if (!flow) break;
    }
  }
  if (!ret) d[u] = -1;
  return ret;
}
int Dinic() {
  int ret = 0;
  for (; bfs();) ret += dfs(S, 23333333);
  return ret;
}
int check(int mid) {
  init();
  S = n * 2 + 1, T = n * 2 + 2;
  for (int i = 1; i <= n; i++) add(S, i, sum[i]), add(i + n, T, 1);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (g[i][j] <= mid) add(i, j + n, 23333333);
  return Dinic() >= k;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &nn, &k);
  for (int i = 1; i <= nn; i++) {
    int x;
    scanf("%d", &x);
    sum[x]++;
  }
  memset(g, 0x3f, sizeof g);
  for (int i = 1; i <= m; i++) {
    int u, v, c;
    scanf("%d%d%d", &u, &v, &c);
    g[v][u] = g[u][v] = min(g[u][v], c);
  }
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        g[i][j] = min(g[i][j], g[i][k] + g[k][j]), g[i][i] = 0;
  int l = -1, r = 2e6;
  while (l + 1 < r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  printf("%d", r == 2e6 ? -1 : r);
  return 0;
}

```