#include <bits/stdc++.h>
using namespace std;
int n, m, s, k, i, sss, top, bot, mid, res, dx, cnt, x, y, z;
vector<pair<int, int> > a[700], b[700];
int c[700], d[700], p[700], match[700];
bool found;
void dijkstra(int t, int x) {
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      q;
  q.push(pair<int, int>(0, x));
  for (int i = 1; i <= n; ++i) d[i] = 1e9;
  d[x] = 0;
  while (q.size()) {
    dx = q.top().first;
    x = q.top().second;
    q.pop();
    if (d[x] != dx) continue;
    b[t].push_back(pair<int, int>(x, dx));
    for (int i = 0; i < a[x].size(); ++i)
      if (d[a[x][i].first] > d[x] + a[x][i].second) {
        d[a[x][i].first] = d[x] + a[x][i].second;
        q.push(pair<int, int>(d[a[x][i].first], a[x][i].first));
      }
  }
}
void dfs(int x) {
  for (int i = 0; i < b[x].size(); ++i) {
    if (b[x][i].second > sss) return;
    int y = b[x][i].first;
    if (c[y] == cnt) continue;
    c[y] = cnt;
    if (match[y] == 0)
      found = true;
    else
      dfs(match[y]);
    if (found) {
      match[y] = x;
      return;
    }
  }
}
bool check(int mid) {
  memset(match, 0, sizeof(match));
  int res = 0;
  sss = mid;
  for (int i = 1; i <= s; ++i) {
    cnt++;
    found = false;
    dfs(i);
    if (found) res++;
  }
  return res >= k;
}
int main() {
  scanf("%d %d %d %d", &n, &m, &s, &k);
  for (i = 1; i <= s; ++i) scanf("%d", &p[i]);
  for (i = 1; i <= m; ++i) {
    scanf("%d %d %d", &x, &y, &z);
    a[x].push_back(pair<int, int>(y, z));
    a[y].push_back(pair<int, int>(x, z));
  }
  for (i = 1; i <= s; ++i) dijkstra(i, p[i]);
  bot = 0;
  top = 1731311;
  res = -1;
  memset(c, 0, sizeof(c));
  cnt = 0;
  while (bot <= top) {
    mid = (bot + top) / 2;
    if (check(mid)) {
      res = mid;
      top = mid - 1;
    } else
      bot = mid + 1;
  }
  printf("%d", res);
}

### Prompt

Create a solution in cpp for 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;
int n, m, s, k, i, sss, top, bot, mid, res, dx, cnt, x, y, z;
vector<pair<int, int> > a[700], b[700];
int c[700], d[700], p[700], match[700];
bool found;
void dijkstra(int t, int x) {
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      q;
  q.push(pair<int, int>(0, x));
  for (int i = 1; i <= n; ++i) d[i] = 1e9;
  d[x] = 0;
  while (q.size()) {
    dx = q.top().first;
    x = q.top().second;
    q.pop();
    if (d[x] != dx) continue;
    b[t].push_back(pair<int, int>(x, dx));
    for (int i = 0; i < a[x].size(); ++i)
      if (d[a[x][i].first] > d[x] + a[x][i].second) {
        d[a[x][i].first] = d[x] + a[x][i].second;
        q.push(pair<int, int>(d[a[x][i].first], a[x][i].first));
      }
  }
}
void dfs(int x) {
  for (int i = 0; i < b[x].size(); ++i) {
    if (b[x][i].second > sss) return;
    int y = b[x][i].first;
    if (c[y] == cnt) continue;
    c[y] = cnt;
    if (match[y] == 0)
      found = true;
    else
      dfs(match[y]);
    if (found) {
      match[y] = x;
      return;
    }
  }
}
bool check(int mid) {
  memset(match, 0, sizeof(match));
  int res = 0;
  sss = mid;
  for (int i = 1; i <= s; ++i) {
    cnt++;
    found = false;
    dfs(i);
    if (found) res++;
  }
  return res >= k;
}
int main() {
  scanf("%d %d %d %d", &n, &m, &s, &k);
  for (i = 1; i <= s; ++i) scanf("%d", &p[i]);
  for (i = 1; i <= m; ++i) {
    scanf("%d %d %d", &x, &y, &z);
    a[x].push_back(pair<int, int>(y, z));
    a[y].push_back(pair<int, int>(x, z));
  }
  for (i = 1; i <= s; ++i) dijkstra(i, p[i]);
  bot = 0;
  top = 1731311;
  res = -1;
  memset(c, 0, sizeof(c));
  cnt = 0;
  while (bot <= top) {
    mid = (bot + top) / 2;
    if (check(mid)) {
      res = mid;
      top = mid - 1;
    } else
      bot = mid + 1;
  }
  printf("%d", res);
}

```