#include <bits/stdc++.h>
inline int rd() {
  int x = 0, p = 1;
  char a = getchar();
  while ((a < 48 || a > 57) && a != '-') a = getchar();
  if (a == '-') p = -p, a = getchar();
  while (a > 47 && a < 58) {
    x = (x << 1) + (x << 3) + (a & 15);
    a = getchar();
  }
  return x * p;
}
inline int max(int x, int y) { return x > y ? x : y; }
inline int min(int x, int y) { return x < y ? x : y; }
const int N = 300002;
struct Edge {
  int to, next;
} edge[N << 1];
int head[N << 1], cnt;
int n, m, k, p;
inline void add(int f, int t) {
  edge[++cnt].next = head[f];
  edge[cnt].to = t;
  head[f] = cnt;
}
int x[N], f[602][602], l, r, ans;
int vis[N], pre[N];
inline int dfs(int u) {
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (!vis[v]) {
      vis[v] = 1;
      if (!pre[v] || dfs(pre[v])) {
        pre[v] = u;
        return 1;
      }
    }
  }
  return 0;
}
inline int check(int mid) {
  cnt = 0;
  memset(head, 0, sizeof head);
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= n; j++)
      if (f[x[i]][j] <= mid) add(i, j);
  int ans = 0;
  memset(pre, 0, sizeof pre);
  for (int i = 1; i <= n; i++) ans += dfs(i), memset(vis, 0, sizeof vis);
  return ans >= p;
}
int main() {
  n = rd(), m = rd(), k = rd(), p = rd();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) f[i][j] = (i ^ j) ? 1000000000 : 0;
  for (int i = 1; i <= k; i++) x[i] = rd();
  for (int i = 1; i <= m; i++) {
    int u = rd(), v = rd(), w = rd();
    f[u][v] = f[v][u] = min(f[v][u], w);
  }
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
  l = 0, r = 1800000;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      r = mid;
    else
      l = mid + 1;
  }
  if (l > 1731311)
    puts("-1");
  else
    printf("%d\n", l);
  return 0;
}

### Prompt

Generate 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>
inline int rd() {
  int x = 0, p = 1;
  char a = getchar();
  while ((a < 48 || a > 57) && a != '-') a = getchar();
  if (a == '-') p = -p, a = getchar();
  while (a > 47 && a < 58) {
    x = (x << 1) + (x << 3) + (a & 15);
    a = getchar();
  }
  return x * p;
}
inline int max(int x, int y) { return x > y ? x : y; }
inline int min(int x, int y) { return x < y ? x : y; }
const int N = 300002;
struct Edge {
  int to, next;
} edge[N << 1];
int head[N << 1], cnt;
int n, m, k, p;
inline void add(int f, int t) {
  edge[++cnt].next = head[f];
  edge[cnt].to = t;
  head[f] = cnt;
}
int x[N], f[602][602], l, r, ans;
int vis[N], pre[N];
inline int dfs(int u) {
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (!vis[v]) {
      vis[v] = 1;
      if (!pre[v] || dfs(pre[v])) {
        pre[v] = u;
        return 1;
      }
    }
  }
  return 0;
}
inline int check(int mid) {
  cnt = 0;
  memset(head, 0, sizeof head);
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= n; j++)
      if (f[x[i]][j] <= mid) add(i, j);
  int ans = 0;
  memset(pre, 0, sizeof pre);
  for (int i = 1; i <= n; i++) ans += dfs(i), memset(vis, 0, sizeof vis);
  return ans >= p;
}
int main() {
  n = rd(), m = rd(), k = rd(), p = rd();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) f[i][j] = (i ^ j) ? 1000000000 : 0;
  for (int i = 1; i <= k; i++) x[i] = rd();
  for (int i = 1; i <= m; i++) {
    int u = rd(), v = rd(), w = rd();
    f[u][v] = f[v][u] = min(f[v][u], w);
  }
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
  l = 0, r = 1800000;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      r = mid;
    else
      l = mid + 1;
  }
  if (l > 1731311)
    puts("-1");
  else
    printf("%d\n", l);
  return 0;
}

```