#include <bits/stdc++.h>
using namespace std;
const int N = 600 + 15;
const int inf = 1e9 + 7;
int v, e, n, k;
int dis[N][N], pos[N], match[N], mp[N][N], vis[N];
inline int read() {
  char ch = getchar();
  int s = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = (s << 3) + (s << 1) + ch - '0';
    ch = getchar();
  }
  return s * f;
}
void floyd() {
  for (int k = 1; k <= v; k++)
    for (int i = 1; i <= v; i++)
      for (int j = 1; j <= v; j++)
        dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
}
bool dfs(int x) {
  for (int i = 1; i <= v; i++) {
    if (vis[i] || !mp[x][i]) continue;
    vis[i] = 1;
    if (match[i] == -1 || dfs(match[i])) {
      match[i] = x;
      return true;
    }
  }
  return false;
}
int check(int l) {
  memset(match, -1, sizeof(match));
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= v; j++) mp[i][j] = (dis[pos[i]][j] <= l);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    memset(vis, 0, sizeof(vis));
    if (dfs(i)) ans++;
  }
  return ans;
}
int main() {
  v = read();
  e = read();
  n = read();
  k = read();
  for (int i = 1; i <= v; i++)
    for (int j = 1; j <= v; j++)
      if (i != j) dis[i][j] = inf;
  for (int i = 1; i <= n; i++) pos[i] = read();
  for (int u, v, w, i = 1; i <= e; i++) {
    u = read();
    v = read();
    w = read();
    dis[u][v] = dis[v][u] = min(dis[u][v], w);
  }
  floyd();
  int l = 0, r = inf, ans = inf;
  while (l < r) {
    int mid = l + r >> 1;
    if (check(mid) >= k)
      r = mid, ans = min(ans, mid);
    else
      l = mid + 1;
  }
  if (ans == inf)
    puts("-1");
  else
    printf("%d\n", ans);
  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;
const int N = 600 + 15;
const int inf = 1e9 + 7;
int v, e, n, k;
int dis[N][N], pos[N], match[N], mp[N][N], vis[N];
inline int read() {
  char ch = getchar();
  int s = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = (s << 3) + (s << 1) + ch - '0';
    ch = getchar();
  }
  return s * f;
}
void floyd() {
  for (int k = 1; k <= v; k++)
    for (int i = 1; i <= v; i++)
      for (int j = 1; j <= v; j++)
        dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
}
bool dfs(int x) {
  for (int i = 1; i <= v; i++) {
    if (vis[i] || !mp[x][i]) continue;
    vis[i] = 1;
    if (match[i] == -1 || dfs(match[i])) {
      match[i] = x;
      return true;
    }
  }
  return false;
}
int check(int l) {
  memset(match, -1, sizeof(match));
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= v; j++) mp[i][j] = (dis[pos[i]][j] <= l);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    memset(vis, 0, sizeof(vis));
    if (dfs(i)) ans++;
  }
  return ans;
}
int main() {
  v = read();
  e = read();
  n = read();
  k = read();
  for (int i = 1; i <= v; i++)
    for (int j = 1; j <= v; j++)
      if (i != j) dis[i][j] = inf;
  for (int i = 1; i <= n; i++) pos[i] = read();
  for (int u, v, w, i = 1; i <= e; i++) {
    u = read();
    v = read();
    w = read();
    dis[u][v] = dis[v][u] = min(dis[u][v], w);
  }
  floyd();
  int l = 0, r = inf, ans = inf;
  while (l < r) {
    int mid = l + r >> 1;
    if (check(mid) >= k)
      r = mid, ans = min(ans, mid);
    else
      l = mid + 1;
  }
  if (ans == inf)
    puts("-1");
  else
    printf("%d\n", ans);
  return 0;
}

```