#include <bits/stdc++.h>
using namespace std;
inline int Read() {
  int x = 0, f = 1, c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int MAXN = 1205;
const int MAXM = 1000005;
const int INF = 0x3f3f3f3f;
namespace Flow {
struct Edge {
  int p, v, w;
} e[MAXM];
int e_cnt, S, T, V, ql, qr, q[MAXN], hed[MAXN], dis[MAXN], cur[MAXN];
inline void Init(int n) {
  V = n;
  for (int i = 0; i < V; i++) hed[i] = 0;
  e_cnt = 1;
}
inline void Addedge(int x, int y, int w) {
  e[++e_cnt] = {y, hed[x], w};
  hed[x] = e_cnt;
  e[++e_cnt] = {x, hed[y], 0};
  hed[y] = e_cnt;
}
inline bool Bfs() {
  for (int i = 0; i < V; i++) dis[i] = 0;
  dis[q[ql = 0] = S] = qr = 1;
  while (ql ^ qr) {
    int x = q[ql++];
    for (int i = hed[x]; i; i = e[i].v)
      if (e[i].w && !dis[e[i].p]) dis[q[qr++] = e[i].p] = dis[x] + 1;
  }
  return dis[T];
}
inline int Dfs(int x, int f) {
  if (x == T) return f;
  int ret = 0, t = 0;
  for (int &i = cur[x]; i; i = e[i].v)
    if (e[i].w && dis[e[i].p] == dis[x] + 1) {
      t = Dfs(e[i].p, min(f, e[i].w));
      f -= t;
      ret += t;
      e[i].w -= t;
      e[i ^ 1].w += t;
      if (!f) return ret;
    }
  return dis[x] = -1, ret;
}
inline int Dinic() {
  int ret = 0;
  while (Bfs()) {
    for (int i = 0; i < V; i++) cur[i] = hed[i];
    ret += Dfs(S, INF);
  }
  return ret;
}
}  // namespace Flow
int n, m, p, q, f[MAXN], dis[MAXN][MAXN];
inline bool Chk(int w) {
  Flow::Init(n + 1 << 1);
  Flow::T = n << 1 | 1;
  for (int i = 1; i <= n; i++) Flow::Addedge(0, i, f[i]);
  for (int i = 1; i <= n; i++) Flow::Addedge(i + n, n << 1 | 1, 1);
  for (int i = 1; i <= n; i++)
    if (f[i])
      for (int j = 1; j <= n; j++)
        if (dis[i][j] <= w) Flow::Addedge(i, j + n, 1);
  return Flow::Dinic() >= q;
}
int main() {
  n = Read(), m = Read(), p = Read(), q = Read();
  for (int i = 1; i <= p; i++) f[Read()]++;
  memset(dis, 0x3f, sizeof dis);
  for (int i = 1; i <= n; i++) dis[i][i] = 0;
  for (int i = 1, x, y, w; i <= m; i++)
    x = Read(), y = Read(), w = Read(),
    dis[x][y] = dis[y][x] = min(dis[x][y], w);
  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, ret = -1;
  while (l <= r) {
    int mid = l + r >> 1;
    if (Chk(mid))
      r = mid - 1, ret = mid;
    else
      l = mid + 1;
  }
  return printf("%d\n", ret), 0;
}

### Prompt

In Cpp, your task is to solve 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;
inline int Read() {
  int x = 0, f = 1, c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int MAXN = 1205;
const int MAXM = 1000005;
const int INF = 0x3f3f3f3f;
namespace Flow {
struct Edge {
  int p, v, w;
} e[MAXM];
int e_cnt, S, T, V, ql, qr, q[MAXN], hed[MAXN], dis[MAXN], cur[MAXN];
inline void Init(int n) {
  V = n;
  for (int i = 0; i < V; i++) hed[i] = 0;
  e_cnt = 1;
}
inline void Addedge(int x, int y, int w) {
  e[++e_cnt] = {y, hed[x], w};
  hed[x] = e_cnt;
  e[++e_cnt] = {x, hed[y], 0};
  hed[y] = e_cnt;
}
inline bool Bfs() {
  for (int i = 0; i < V; i++) dis[i] = 0;
  dis[q[ql = 0] = S] = qr = 1;
  while (ql ^ qr) {
    int x = q[ql++];
    for (int i = hed[x]; i; i = e[i].v)
      if (e[i].w && !dis[e[i].p]) dis[q[qr++] = e[i].p] = dis[x] + 1;
  }
  return dis[T];
}
inline int Dfs(int x, int f) {
  if (x == T) return f;
  int ret = 0, t = 0;
  for (int &i = cur[x]; i; i = e[i].v)
    if (e[i].w && dis[e[i].p] == dis[x] + 1) {
      t = Dfs(e[i].p, min(f, e[i].w));
      f -= t;
      ret += t;
      e[i].w -= t;
      e[i ^ 1].w += t;
      if (!f) return ret;
    }
  return dis[x] = -1, ret;
}
inline int Dinic() {
  int ret = 0;
  while (Bfs()) {
    for (int i = 0; i < V; i++) cur[i] = hed[i];
    ret += Dfs(S, INF);
  }
  return ret;
}
}  // namespace Flow
int n, m, p, q, f[MAXN], dis[MAXN][MAXN];
inline bool Chk(int w) {
  Flow::Init(n + 1 << 1);
  Flow::T = n << 1 | 1;
  for (int i = 1; i <= n; i++) Flow::Addedge(0, i, f[i]);
  for (int i = 1; i <= n; i++) Flow::Addedge(i + n, n << 1 | 1, 1);
  for (int i = 1; i <= n; i++)
    if (f[i])
      for (int j = 1; j <= n; j++)
        if (dis[i][j] <= w) Flow::Addedge(i, j + n, 1);
  return Flow::Dinic() >= q;
}
int main() {
  n = Read(), m = Read(), p = Read(), q = Read();
  for (int i = 1; i <= p; i++) f[Read()]++;
  memset(dis, 0x3f, sizeof dis);
  for (int i = 1; i <= n; i++) dis[i][i] = 0;
  for (int i = 1, x, y, w; i <= m; i++)
    x = Read(), y = Read(), w = Read(),
    dis[x][y] = dis[y][x] = min(dis[x][y], w);
  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, ret = -1;
  while (l <= r) {
    int mid = l + r >> 1;
    if (Chk(mid))
      r = mid - 1, ret = mid;
    else
      l = mid + 1;
  }
  return printf("%d\n", ret), 0;
}

```