#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1210, MAXM = 20010;
int n, m, k, t, dis[MAXN][MAXN], a[MAXN], match[MAXN];
bool vis[MAXN], con[MAXN][MAXN];
struct Edge {
  int ver, nxt, w;
} e[MAXM];
int head[MAXN], tot = 0;
void add(int u, int v, int w) {
  e[++tot].ver = v;
  e[tot].w = w;
  e[tot].nxt = head[u];
  head[u] = tot;
}
struct Que {
  int u, d;
  bool operator<(const Que &f) const { return d > f.d; }
};
priority_queue<Que> q;
void Dijkstra(int s) {
  int i;
  dis[s][s] = 0;
  q.push((Que){s, 0});
  while (!q.empty()) {
    Que t = q.top();
    q.pop();
    int u = t.u, d = t.d;
    if (vis[u]) continue;
    vis[u] = true;
    for (i = head[u]; i; i = e[i].nxt) {
      int v = e[i].ver, w = e[i].w;
      if (dis[s][v] > dis[s][u] + w) {
        dis[s][v] = dis[s][u] + w;
        q.push((Que){v, dis[s][v]});
      }
    }
  }
  return;
}
bool dfs(int u) {
  int v;
  for (v = 1; v <= n; v++) {
    if (con[u][v] && !vis[v]) {
      vis[v] = 1;
      if (!match[v] || dfs(match[v])) {
        match[v] = u;
        return 1;
      }
    }
  }
  return 0;
}
int hungary() {
  memset(match, 0, sizeof(match));
  int res = 0, i;
  for (i = 1; i <= k; i++) {
    memset(vis, false, sizeof(vis));
    if (dfs(i)) res++;
  }
  return res;
}
bool check(int mid) {
  memset(con, false, sizeof(con));
  int i, j;
  for (i = 1; i <= k; i++)
    for (j = 1; j <= n; j++)
      if (dis[a[i]][j] <= mid) con[i][j] = true;
  int res = hungary();
  return res >= t;
}
int main() {
  int i, j;
  scanf("%d %d %d %d", &n, &m, &k, &t);
  for (i = 1; i <= k; i++) scanf("%d", &a[i]);
  memset(dis, 0x3F, sizeof(dis));
  for (i = 1; i <= m; i++) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    add(u, v, w);
    add(v, u, w);
  }
  memset(dis, 0x3F, sizeof(dis));
  for (i = 1; i <= n; i++) {
    memset(vis, false, sizeof(vis));
    Dijkstra(i);
  }
  int l = 0, r = 1731312, mid, res = -1;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (check(mid)) {
      res = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  printf("%d", res);
  return 0;
}

### Prompt

Please formulate 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;
const int MAXN = 1210, MAXM = 20010;
int n, m, k, t, dis[MAXN][MAXN], a[MAXN], match[MAXN];
bool vis[MAXN], con[MAXN][MAXN];
struct Edge {
  int ver, nxt, w;
} e[MAXM];
int head[MAXN], tot = 0;
void add(int u, int v, int w) {
  e[++tot].ver = v;
  e[tot].w = w;
  e[tot].nxt = head[u];
  head[u] = tot;
}
struct Que {
  int u, d;
  bool operator<(const Que &f) const { return d > f.d; }
};
priority_queue<Que> q;
void Dijkstra(int s) {
  int i;
  dis[s][s] = 0;
  q.push((Que){s, 0});
  while (!q.empty()) {
    Que t = q.top();
    q.pop();
    int u = t.u, d = t.d;
    if (vis[u]) continue;
    vis[u] = true;
    for (i = head[u]; i; i = e[i].nxt) {
      int v = e[i].ver, w = e[i].w;
      if (dis[s][v] > dis[s][u] + w) {
        dis[s][v] = dis[s][u] + w;
        q.push((Que){v, dis[s][v]});
      }
    }
  }
  return;
}
bool dfs(int u) {
  int v;
  for (v = 1; v <= n; v++) {
    if (con[u][v] && !vis[v]) {
      vis[v] = 1;
      if (!match[v] || dfs(match[v])) {
        match[v] = u;
        return 1;
      }
    }
  }
  return 0;
}
int hungary() {
  memset(match, 0, sizeof(match));
  int res = 0, i;
  for (i = 1; i <= k; i++) {
    memset(vis, false, sizeof(vis));
    if (dfs(i)) res++;
  }
  return res;
}
bool check(int mid) {
  memset(con, false, sizeof(con));
  int i, j;
  for (i = 1; i <= k; i++)
    for (j = 1; j <= n; j++)
      if (dis[a[i]][j] <= mid) con[i][j] = true;
  int res = hungary();
  return res >= t;
}
int main() {
  int i, j;
  scanf("%d %d %d %d", &n, &m, &k, &t);
  for (i = 1; i <= k; i++) scanf("%d", &a[i]);
  memset(dis, 0x3F, sizeof(dis));
  for (i = 1; i <= m; i++) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    add(u, v, w);
    add(v, u, w);
  }
  memset(dis, 0x3F, sizeof(dis));
  for (i = 1; i <= n; i++) {
    memset(vis, false, sizeof(vis));
    Dijkstra(i);
  }
  int l = 0, r = 1731312, mid, res = -1;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (check(mid)) {
      res = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  printf("%d", res);
  return 0;
}

```