#include <bits/stdc++.h>
using namespace std;
const int inf = 1731312;
struct node {
  int v, next, c;
} eg[20005 * 2];
int head[605], tot, p[202], ji[605], d[605][605], v, match[605], n;
bool a[605][605];
void add(int x, int y, int c) {
  eg[tot] = node{y, head[x], c};
  head[x] = tot++;
}
void dist(int src) {
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      Q;
  Q.push(pair<int, int>(0, src));
  while (!Q.empty()) {
    int p = Q.top().second;
    int dis = Q.top().first;
    Q.pop();
    if (d[src][p] < dis) continue;
    d[src][p] = dis;
    for (int i = head[p]; ~i; i = eg[i].next) {
      int v = eg[i].v;
      if (d[src][v] < dis + eg[i].c) continue;
      d[src][v] = dis + eg[i].c;
      Q.push(pair<int, int>(d[src][v], v));
    }
  }
}
bool dfs(int p) {
  for (int i = 1; i <= v; i++) {
    if (ji[i] || !a[p][i]) continue;
    ji[i] = 1;
    if (match[i] == -1 || dfs(match[i])) {
      match[i] = p;
      return 1;
    }
  }
  return 0;
}
int check(int x) {
  memset(match, -1, sizeof(match));
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= v; j++) a[i][j] = d[p[i]][j] <= x;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    memset(ji, 0, sizeof(ji));
    if (dfs(i)) ans++;
  }
  return ans;
}
int main() {
  int e, k;
  scanf("%d%d%d%d", &v, &e, &n, &k);
  tot = 0;
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= v; i++)
    for (int j = 1; j <= v; j++) d[i][j] = inf;
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 0; i < e; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    add(x, y, z);
    add(y, x, z);
  }
  for (int i = 1; i <= n; i++) {
    if (!ji[p[i]]) {
      ji[p[i]] = 1;
      dist(p[i]);
    }
  }
  int l = 0, r = inf, ans = inf;
  while (l <= r) {
    int mid = l + r >> 1;
    if (check(mid) >= k)
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  if (ans == inf) ans = -1;
  printf("%d\n", ans);
  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>
using namespace std;
const int inf = 1731312;
struct node {
  int v, next, c;
} eg[20005 * 2];
int head[605], tot, p[202], ji[605], d[605][605], v, match[605], n;
bool a[605][605];
void add(int x, int y, int c) {
  eg[tot] = node{y, head[x], c};
  head[x] = tot++;
}
void dist(int src) {
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      Q;
  Q.push(pair<int, int>(0, src));
  while (!Q.empty()) {
    int p = Q.top().second;
    int dis = Q.top().first;
    Q.pop();
    if (d[src][p] < dis) continue;
    d[src][p] = dis;
    for (int i = head[p]; ~i; i = eg[i].next) {
      int v = eg[i].v;
      if (d[src][v] < dis + eg[i].c) continue;
      d[src][v] = dis + eg[i].c;
      Q.push(pair<int, int>(d[src][v], v));
    }
  }
}
bool dfs(int p) {
  for (int i = 1; i <= v; i++) {
    if (ji[i] || !a[p][i]) continue;
    ji[i] = 1;
    if (match[i] == -1 || dfs(match[i])) {
      match[i] = p;
      return 1;
    }
  }
  return 0;
}
int check(int x) {
  memset(match, -1, sizeof(match));
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= v; j++) a[i][j] = d[p[i]][j] <= x;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    memset(ji, 0, sizeof(ji));
    if (dfs(i)) ans++;
  }
  return ans;
}
int main() {
  int e, k;
  scanf("%d%d%d%d", &v, &e, &n, &k);
  tot = 0;
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= v; i++)
    for (int j = 1; j <= v; j++) d[i][j] = inf;
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 0; i < e; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    add(x, y, z);
    add(y, x, z);
  }
  for (int i = 1; i <= n; i++) {
    if (!ji[p[i]]) {
      ji[p[i]] = 1;
      dist(p[i]);
    }
  }
  int l = 0, r = inf, ans = inf;
  while (l <= r) {
    int mid = l + r >> 1;
    if (check(mid) >= k)
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  if (ans == inf) ans = -1;
  printf("%d\n", ans);
  return 0;
}

```