#include <bits/stdc++.h>
using namespace std;
inline int read(int f = 1, int x = 0, char ch = ' ') {
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = -1;
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f * x;
}
const int N = 6e2 + 5, M = 4e4 + 5;
struct Edge {
  int next, to, w;
  Edge(int next = 0, int to = 0, int w = 0) : next(next), to(to), w(w){};
} edge[M];
int tot, head[N];
void _add(int x, int y, int z) {
  edge[++tot] = Edge(head[x], y, z), head[x] = tot;
}
void add(int x, int y, int z) { _add(x, y, z), _add(y, x, z); }
int n, m, k, q, l, r, f[N][N], c[N], s[N], ans = -1, match[N], vis[N];
void dijkstra(int s, int *d) {
  memset(d + 1, 0x3f, sizeof(int) * n);
  priority_queue<pair<int, int> > q;
  q.push(make_pair(d[s] = 0, s));
  while (q.size()) {
    int x = q.top().second, dx = -q.top().first;
    q.pop();
    if (dx != d[x]) continue;
    for (int i = head[x]; i; i = edge[i].next) {
      int y = edge[i].to, z = edge[i].w;
      if (d[y] > d[x] + z) d[y] = d[x] + z, q.push(make_pair(-d[y], y));
    }
  }
}
bool dfs(int x, int mid) {
  for (int y = 1; y <= n; ++y)
    if (!vis[y] && f[x][y] <= mid) {
      vis[y] = 1;
      if (match[y] == -1 || dfs(match[y], mid)) return match[y] = x, true;
    }
  return false;
}
bool valid(int mid) {
  int cnt = 0;
  memset(match + 1, -1, sizeof(int) * n);
  for (int i = 1; i <= k; ++i) {
    memset(vis, 0, sizeof(vis));
    if ((cnt += dfs(i, mid)) >= q) return true;
  }
  return false;
}
int main() {
  n = read(), m = read(), k = read(), q = read();
  for (int i = 1; i <= k; ++i) s[i] = read();
  for (int i = 1; i <= m; ++i) {
    int x = read(), y = read(), z = read();
    add(x, y, z), r += z;
  }
  for (int i = 1; i <= k; ++i) dijkstra(s[i], f[i]);
  r = min(1731311, r);
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (valid(mid))
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  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;
inline int read(int f = 1, int x = 0, char ch = ' ') {
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = -1;
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f * x;
}
const int N = 6e2 + 5, M = 4e4 + 5;
struct Edge {
  int next, to, w;
  Edge(int next = 0, int to = 0, int w = 0) : next(next), to(to), w(w){};
} edge[M];
int tot, head[N];
void _add(int x, int y, int z) {
  edge[++tot] = Edge(head[x], y, z), head[x] = tot;
}
void add(int x, int y, int z) { _add(x, y, z), _add(y, x, z); }
int n, m, k, q, l, r, f[N][N], c[N], s[N], ans = -1, match[N], vis[N];
void dijkstra(int s, int *d) {
  memset(d + 1, 0x3f, sizeof(int) * n);
  priority_queue<pair<int, int> > q;
  q.push(make_pair(d[s] = 0, s));
  while (q.size()) {
    int x = q.top().second, dx = -q.top().first;
    q.pop();
    if (dx != d[x]) continue;
    for (int i = head[x]; i; i = edge[i].next) {
      int y = edge[i].to, z = edge[i].w;
      if (d[y] > d[x] + z) d[y] = d[x] + z, q.push(make_pair(-d[y], y));
    }
  }
}
bool dfs(int x, int mid) {
  for (int y = 1; y <= n; ++y)
    if (!vis[y] && f[x][y] <= mid) {
      vis[y] = 1;
      if (match[y] == -1 || dfs(match[y], mid)) return match[y] = x, true;
    }
  return false;
}
bool valid(int mid) {
  int cnt = 0;
  memset(match + 1, -1, sizeof(int) * n);
  for (int i = 1; i <= k; ++i) {
    memset(vis, 0, sizeof(vis));
    if ((cnt += dfs(i, mid)) >= q) return true;
  }
  return false;
}
int main() {
  n = read(), m = read(), k = read(), q = read();
  for (int i = 1; i <= k; ++i) s[i] = read();
  for (int i = 1; i <= m; ++i) {
    int x = read(), y = read(), z = read();
    add(x, y, z), r += z;
  }
  for (int i = 1; i <= k; ++i) dijkstra(s[i], f[i]);
  r = min(1731311, r);
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (valid(mid))
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%d\n", ans);
  return 0;
}

```