#include <bits/stdc++.h>
using namespace std;
inline int QRead() {
  int acur = 0;
  char ch = getchar();
  while ('0' > ch || ch > '9') ch = getchar();
  while ('0' <= ch && ch <= '9')
    acur = (acur << 3) + (acur << 1) + ch - '0', ch = getchar();
  return acur;
}
const int NTEM = 200, NPNT = 600;
struct EDGE {
  int to, val;
};
vector<EDGE> lnk[NTEM + NPNT + 3];
int frm[NTEM + 3], dis[NPNT + 3][NPNT + 3], vis[NTEM + NPNT + 3],
    mat[NTEM + NPNT + 3];
int npnt, nedg, ntem, nreq, maxdis, lim, vistag;
bool DFS(int u) {
  vis[u] = vistag;
  for (auto it : lnk[u]) {
    int val = it.val, v = it.to;
    if (vis[v] == vistag || val > lim) continue;
    vis[v] = vistag;
    if (!mat[v] || DFS(mat[v])) {
      mat[u] = v;
      mat[v] = u;
      return true;
    }
  }
  return false;
}
bool Match(int mid) {
  lim = mid;
  int cur = 0;
  memset(mat, 0, sizeof mat);
  for (int i = 1; i <= ntem; i++)
    if (!mat[i]) {
      vistag++;
      cur += DFS(i);
    }
  return cur >= nreq;
}
void AddEdge(int u, int v, int val) {
  lnk[u].push_back((EDGE){v, val});
  lnk[v].push_back((EDGE){u, val});
}
int BinarySearch() {
  int Lef = 0, Rig = maxdis + 1;
  while (Lef + 1 < Rig) {
    int Mid = (Lef + Rig) >> 1;
    if (Match(Mid))
      Rig = Mid;
    else
      Lef = Mid;
  }
  return Rig;
}
int main() {
  memset(dis, 0x3f, sizeof dis);
  int inf = dis[0][0];
  npnt = QRead();
  nedg = QRead();
  ntem = QRead();
  nreq = QRead();
  for (int i = 1; i <= ntem; i++) frm[i] = QRead();
  for (int i = 1; i <= nedg; i++) {
    int u, v, far;
    u = QRead();
    v = QRead();
    far = QRead();
    dis[u][v] = dis[v][u] = min(dis[u][v], far);
  }
  for (int i = 1; i <= npnt; i++) dis[i][i] = 0;
  for (int k = 1; k <= npnt; k++)
    for (int i = 1; i <= npnt; i++)
      for (int j = 1; j <= npnt; j++) {
        if (i == k || j == k || i == j) continue;
        if (dis[i][k] == inf || dis[k][j] == inf) continue;
        dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
      }
  for (int i = 1; i <= npnt; i++)
    for (int j = 1; j <= npnt; j++)
      if (dis[i][j] != inf) maxdis = max(maxdis, dis[i][j]);
  for (int i = 1; i <= ntem; i++)
    for (int j = 1; j <= npnt; j++) AddEdge(i, j + ntem, dis[frm[i]][j]);
  if (Match(0)) {
    printf("%d\n", 0);
    return 0;
  }
  int res = BinarySearch();
  if (res <= maxdis && Match(res))
    printf("%d\n", res);
  else
    printf("-1\n");
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
inline int QRead() {
  int acur = 0;
  char ch = getchar();
  while ('0' > ch || ch > '9') ch = getchar();
  while ('0' <= ch && ch <= '9')
    acur = (acur << 3) + (acur << 1) + ch - '0', ch = getchar();
  return acur;
}
const int NTEM = 200, NPNT = 600;
struct EDGE {
  int to, val;
};
vector<EDGE> lnk[NTEM + NPNT + 3];
int frm[NTEM + 3], dis[NPNT + 3][NPNT + 3], vis[NTEM + NPNT + 3],
    mat[NTEM + NPNT + 3];
int npnt, nedg, ntem, nreq, maxdis, lim, vistag;
bool DFS(int u) {
  vis[u] = vistag;
  for (auto it : lnk[u]) {
    int val = it.val, v = it.to;
    if (vis[v] == vistag || val > lim) continue;
    vis[v] = vistag;
    if (!mat[v] || DFS(mat[v])) {
      mat[u] = v;
      mat[v] = u;
      return true;
    }
  }
  return false;
}
bool Match(int mid) {
  lim = mid;
  int cur = 0;
  memset(mat, 0, sizeof mat);
  for (int i = 1; i <= ntem; i++)
    if (!mat[i]) {
      vistag++;
      cur += DFS(i);
    }
  return cur >= nreq;
}
void AddEdge(int u, int v, int val) {
  lnk[u].push_back((EDGE){v, val});
  lnk[v].push_back((EDGE){u, val});
}
int BinarySearch() {
  int Lef = 0, Rig = maxdis + 1;
  while (Lef + 1 < Rig) {
    int Mid = (Lef + Rig) >> 1;
    if (Match(Mid))
      Rig = Mid;
    else
      Lef = Mid;
  }
  return Rig;
}
int main() {
  memset(dis, 0x3f, sizeof dis);
  int inf = dis[0][0];
  npnt = QRead();
  nedg = QRead();
  ntem = QRead();
  nreq = QRead();
  for (int i = 1; i <= ntem; i++) frm[i] = QRead();
  for (int i = 1; i <= nedg; i++) {
    int u, v, far;
    u = QRead();
    v = QRead();
    far = QRead();
    dis[u][v] = dis[v][u] = min(dis[u][v], far);
  }
  for (int i = 1; i <= npnt; i++) dis[i][i] = 0;
  for (int k = 1; k <= npnt; k++)
    for (int i = 1; i <= npnt; i++)
      for (int j = 1; j <= npnt; j++) {
        if (i == k || j == k || i == j) continue;
        if (dis[i][k] == inf || dis[k][j] == inf) continue;
        dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
      }
  for (int i = 1; i <= npnt; i++)
    for (int j = 1; j <= npnt; j++)
      if (dis[i][j] != inf) maxdis = max(maxdis, dis[i][j]);
  for (int i = 1; i <= ntem; i++)
    for (int j = 1; j <= npnt; j++) AddEdge(i, j + ntem, dis[frm[i]][j]);
  if (Match(0)) {
    printf("%d\n", 0);
    return 0;
  }
  int res = BinarySearch();
  if (res <= maxdis && Match(res))
    printf("%d\n", res);
  else
    printf("-1\n");
  return 0;
}

```