#include <bits/stdc++.h>
using namespace std;
const int N = 610, M = 20010, INF = 1e9, UP = 1731311;
int n, m, nn, kk, xx, yy, zz;
int w[N][N], col[N], dis[N][N], link[N], usd[N];
int dfs(int x) {
  for (int i = 1; i <= n; ++i)
    if (w[x][i] && !usd[i]) {
      usd[i] = 1;
      if (!link[i] || dfs(link[i])) {
        link[i] = x;
        return 1;
      }
    }
  return 0;
}
int check() {
  int ans = 0;
  memset(link, 0, sizeof link);
  for (int i = 1; i <= nn; ++i) {
    memset(usd, 0, sizeof usd);
    if (dfs(i)) ans++;
  }
  return ans;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &nn, &kk);
  for (int i = 1; i <= nn; ++i) scanf("%d", &col[i]);
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= n; ++j) dis[i][j] = INF;
  for (int i = 1; i <= n; ++i) dis[i][i] = 0;
  for (int i = 1; i <= m; ++i)
    scanf("%d%d%d", &xx, &yy, &zz),
        dis[xx][yy] = dis[yy][xx] = min(dis[xx][yy], zz);
  for (int k = 1; k <= n; ++k)
    for (int i = 1; i <= n; ++i)
      if (i != k)
        for (int j = 1; j <= n; ++j)
          if (j != i && j != k)
            if (dis[i][j] > dis[i][k] + dis[k][j])
              dis[i][j] = dis[i][k] + dis[k][j];
  int ans = -1, l = 0, r = UP;
  while (l <= r) {
    int mid = (l + r) >> 1;
    for (int i = 1; i <= nn; ++i)
      for (int j = 1; j <= n; ++j) w[i][j] = (dis[col[i]][j] <= mid ? 1 : 0);
    if (check() >= kk)
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%d\n", ans);
  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;
const int N = 610, M = 20010, INF = 1e9, UP = 1731311;
int n, m, nn, kk, xx, yy, zz;
int w[N][N], col[N], dis[N][N], link[N], usd[N];
int dfs(int x) {
  for (int i = 1; i <= n; ++i)
    if (w[x][i] && !usd[i]) {
      usd[i] = 1;
      if (!link[i] || dfs(link[i])) {
        link[i] = x;
        return 1;
      }
    }
  return 0;
}
int check() {
  int ans = 0;
  memset(link, 0, sizeof link);
  for (int i = 1; i <= nn; ++i) {
    memset(usd, 0, sizeof usd);
    if (dfs(i)) ans++;
  }
  return ans;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &nn, &kk);
  for (int i = 1; i <= nn; ++i) scanf("%d", &col[i]);
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= n; ++j) dis[i][j] = INF;
  for (int i = 1; i <= n; ++i) dis[i][i] = 0;
  for (int i = 1; i <= m; ++i)
    scanf("%d%d%d", &xx, &yy, &zz),
        dis[xx][yy] = dis[yy][xx] = min(dis[xx][yy], zz);
  for (int k = 1; k <= n; ++k)
    for (int i = 1; i <= n; ++i)
      if (i != k)
        for (int j = 1; j <= n; ++j)
          if (j != i && j != k)
            if (dis[i][j] > dis[i][k] + dis[k][j])
              dis[i][j] = dis[i][k] + dis[k][j];
  int ans = -1, l = 0, r = UP;
  while (l <= r) {
    int mid = (l + r) >> 1;
    for (int i = 1; i <= nn; ++i)
      for (int j = 1; j <= n; ++j) w[i][j] = (dis[col[i]][j] <= mid ? 1 : 0);
    if (check() >= kk)
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%d\n", ans);
  return 0;
}

```