#include <bits/stdc++.h>
using namespace std;
bool G[605][605];
int n, m, x, y, z, M, K, ans, c[605], f[605][605], use[605], match[605];
inline bool dfs(int u) {
  for (int v = 1; v <= n; v++) {
    if (!G[u][v]) continue;
    if (use[v]) continue;
    use[v] = 1;
    if (!match[v] || dfs(match[v])) {
      match[v] = u;
      return 1;
    }
  }
  return 0;
}
inline bool check(int x) {
  for (int i = 1; i <= M; i++)
    for (int j = 1; j <= n; j++) {
      if (f[c[i]][j] <= x)
        G[i][j] = 1;
      else
        G[i][j] = 0;
    }
  ans = 0;
  memset(match, 0, sizeof(match));
  for (int i = 1; i <= M; i++) {
    memset(use, 0, sizeof(use));
    if (dfs(i)) ans++;
  }
  return ans >= K;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &M, &K);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) f[i][j] = 1e9;
  for (int i = 1; i <= M; i++) scanf("%d", &c[i]);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &x, &y, &z);
    f[x][y] = f[y][x] = min(f[x][y], z);
  }
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
  for (int i = 1; i <= n; i++) f[i][i] = 0;
  int l = 0;
  int r = 1731311;
  if (!check(r)) return puts("-1"), 0;
  while (l + 1 < r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  if (check(l))
    printf("%d\n", l);
  else
    printf("%d\n", r);
  return 0;
}

### Prompt

Generate 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;
bool G[605][605];
int n, m, x, y, z, M, K, ans, c[605], f[605][605], use[605], match[605];
inline bool dfs(int u) {
  for (int v = 1; v <= n; v++) {
    if (!G[u][v]) continue;
    if (use[v]) continue;
    use[v] = 1;
    if (!match[v] || dfs(match[v])) {
      match[v] = u;
      return 1;
    }
  }
  return 0;
}
inline bool check(int x) {
  for (int i = 1; i <= M; i++)
    for (int j = 1; j <= n; j++) {
      if (f[c[i]][j] <= x)
        G[i][j] = 1;
      else
        G[i][j] = 0;
    }
  ans = 0;
  memset(match, 0, sizeof(match));
  for (int i = 1; i <= M; i++) {
    memset(use, 0, sizeof(use));
    if (dfs(i)) ans++;
  }
  return ans >= K;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &M, &K);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) f[i][j] = 1e9;
  for (int i = 1; i <= M; i++) scanf("%d", &c[i]);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &x, &y, &z);
    f[x][y] = f[y][x] = min(f[x][y], z);
  }
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
  for (int i = 1; i <= n; i++) f[i][i] = 0;
  int l = 0;
  int r = 1731311;
  if (!check(r)) return puts("-1"), 0;
  while (l + 1 < r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  if (check(l))
    printf("%d\n", l);
  else
    printf("%d\n", r);
  return 0;
}

```