#include <bits/stdc++.h>
using namespace std;
int n, m, p, r, dis[610][610], occ[610];
namespace MaxFlow {
const int N = 10000, M = 2000000;
int head[N], cur[N], dep[N], cnt, S, T, res;
struct node {
  int to, next, val;
} edge[M];
void ae(int u, int v, int w) {
  edge[cnt].next = head[u], edge[cnt].to = v, edge[cnt].val = w,
  head[u] = cnt++;
  edge[cnt].next = head[v], edge[cnt].to = u, edge[cnt].val = 0,
  head[v] = cnt++;
}
queue<int> q;
inline bool bfs() {
  memset(dep, 0, sizeof(dep)), q.push(S), dep[S] = 1;
  while (!q.empty()) {
    register int x = q.front();
    q.pop();
    for (register int i = cur[x] = head[x]; i != -1; i = edge[i].next)
      if (edge[i].val && !dep[edge[i].to])
        dep[edge[i].to] = dep[x] + 1, q.push(edge[i].to);
  }
  return dep[T] > 0;
}
bool reach;
inline int dfs(int x, int flow) {
  if (x == T) {
    res += flow;
    reach = true;
    return flow;
  }
  int used = 0;
  for (register int &i = cur[x]; i != -1; i = edge[i].next) {
    if (!edge[i].val || dep[edge[i].to] != dep[x] + 1) continue;
    register int ff = dfs(edge[i].to, min(edge[i].val, flow - used));
    if (ff) {
      edge[i].val -= ff;
      edge[i ^ 1].val += ff;
      used += ff;
      if (used == flow) break;
    }
  }
  return used;
}
inline void Dinic() {
  while (bfs()) {
    reach = true;
    while (reach) reach = false, dfs(S, 0x3f3f3f3f);
  }
}
}  // namespace MaxFlow
using namespace MaxFlow;
int che(int ip) {
  memset(head, -1, sizeof(head)), cnt = res = 0;
  for (int i = 1; i <= n; i++) ae(S, i, occ[i]), ae(i + n, T, 1);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (dis[i][j] <= ip) ae(i, j + n, 0x3f3f3f3f);
  Dinic();
  return res >= r;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &p, &r), memset(dis, 0x3f3f3f3f, sizeof(dis)),
      S = n * 2 + 1, T = n * 2 + 2;
  for (int i = 1; i <= n; i++) dis[i][i] = 0;
  for (int i = 1, x; i <= p; i++) scanf("%d", &x), occ[x]++;
  for (int i = 1, x, y, z; i <= m; i++)
    scanf("%d%d%d", &x, &y, &z), dis[x][y] = dis[y][x] = min(dis[x][y], z);
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
  int L = 0, R = 1731311;
  while (L < R) {
    int mid = (L + R) >> 1;
    if (che(mid))
      R = mid;
    else
      L = mid + 1;
  }
  if (!che(R))
    puts("-1");
  else
    printf("%d\n", R);
  return 0;
}

### Prompt

Your task is to create 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;
int n, m, p, r, dis[610][610], occ[610];
namespace MaxFlow {
const int N = 10000, M = 2000000;
int head[N], cur[N], dep[N], cnt, S, T, res;
struct node {
  int to, next, val;
} edge[M];
void ae(int u, int v, int w) {
  edge[cnt].next = head[u], edge[cnt].to = v, edge[cnt].val = w,
  head[u] = cnt++;
  edge[cnt].next = head[v], edge[cnt].to = u, edge[cnt].val = 0,
  head[v] = cnt++;
}
queue<int> q;
inline bool bfs() {
  memset(dep, 0, sizeof(dep)), q.push(S), dep[S] = 1;
  while (!q.empty()) {
    register int x = q.front();
    q.pop();
    for (register int i = cur[x] = head[x]; i != -1; i = edge[i].next)
      if (edge[i].val && !dep[edge[i].to])
        dep[edge[i].to] = dep[x] + 1, q.push(edge[i].to);
  }
  return dep[T] > 0;
}
bool reach;
inline int dfs(int x, int flow) {
  if (x == T) {
    res += flow;
    reach = true;
    return flow;
  }
  int used = 0;
  for (register int &i = cur[x]; i != -1; i = edge[i].next) {
    if (!edge[i].val || dep[edge[i].to] != dep[x] + 1) continue;
    register int ff = dfs(edge[i].to, min(edge[i].val, flow - used));
    if (ff) {
      edge[i].val -= ff;
      edge[i ^ 1].val += ff;
      used += ff;
      if (used == flow) break;
    }
  }
  return used;
}
inline void Dinic() {
  while (bfs()) {
    reach = true;
    while (reach) reach = false, dfs(S, 0x3f3f3f3f);
  }
}
}  // namespace MaxFlow
using namespace MaxFlow;
int che(int ip) {
  memset(head, -1, sizeof(head)), cnt = res = 0;
  for (int i = 1; i <= n; i++) ae(S, i, occ[i]), ae(i + n, T, 1);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (dis[i][j] <= ip) ae(i, j + n, 0x3f3f3f3f);
  Dinic();
  return res >= r;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &p, &r), memset(dis, 0x3f3f3f3f, sizeof(dis)),
      S = n * 2 + 1, T = n * 2 + 2;
  for (int i = 1; i <= n; i++) dis[i][i] = 0;
  for (int i = 1, x; i <= p; i++) scanf("%d", &x), occ[x]++;
  for (int i = 1, x, y, z; i <= m; i++)
    scanf("%d%d%d", &x, &y, &z), dis[x][y] = dis[y][x] = min(dis[x][y], z);
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
  int L = 0, R = 1731311;
  while (L < R) {
    int mid = (L + R) >> 1;
    if (che(mid))
      R = mid;
    else
      L = mid + 1;
  }
  if (!che(R))
    puts("-1");
  else
    printf("%d\n", R);
  return 0;
}

```