#include <bits/stdc++.h>
using namespace std;
namespace WXHAKIOI2018 {
struct edge {
  int r, nxt, w;
} e[700 * 600];
int S, T, head[1100], dep[1100], q[1100], l, r, esz, cur[1100];
void init(int s, int t) {
  esz = 1, S = s, T = t;
  for (int i = S; i <= T; ++i) head[i] = 0;
}
void addedge(int u, int v, int w) {
  e[++esz].r = v;
  e[esz].nxt = head[u];
  head[u] = esz;
  e[esz].w = w;
  e[++esz].r = u;
  e[esz].nxt = head[v];
  head[v] = esz;
  e[esz].w = 0;
}
int bfs() {
  for (int i = S; i <= T; ++i) dep[i] = 0, cur[i] = head[i];
  l = r = 0, dep[q[r++] = S] = 1;
  while (l < r) {
    int u = q[l++];
    for (int t = head[u]; t; t = e[t].nxt)
      if (e[t].w && !dep[e[t].r]) dep[q[r++] = e[t].r] = dep[u] + 1;
  }
  return dep[T] != 0;
}
int find(int u, int flow) {
  if (u == T) return flow;
  int used = 0, a = 0;
  for (int& t = cur[u]; t; t = e[t].nxt)
    if (e[t].w && dep[e[t].r] == dep[u] + 1) {
      a = find(e[t].r, min(flow - used, e[t].w));
      e[t].w -= a, e[t ^ 1].w += a, used += a;
      if (used == flow) return used;
    }
  if (!used) dep[u] = -1;
  return used;
}
int dinic() {
  int ans = 0, a;
  while (bfs())
    while (a = find(S, 1831311)) ans += a;
  return ans;
}
};  // namespace WXHAKIOI2018
int f[1100][1100], v, e, n, k;
int a[1100];
bool check(int ans) {
  WXHAKIOI2018::init(0, n + v + 1);
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= v; ++j)
      if (f[a[i]][j] <= ans) WXHAKIOI2018::addedge(i, j + n, 1831311);
  for (int i = 1; i <= n; ++i) WXHAKIOI2018::addedge(WXHAKIOI2018::S, i, 1);
  for (int i = 1; i <= v; ++i) WXHAKIOI2018::addedge(i + n, WXHAKIOI2018::T, 1);
  return WXHAKIOI2018::dinic() >= k;
}
int main() {
  scanf("%d%d%d%d", &v, &e, &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= v; ++i)
    for (int j = 1; j <= v; ++j)
      if (i != j) f[i][j] = 1831311;
  for (int i = 1, a, b, t; i <= e; ++i) {
    scanf("%d%d%d", &a, &b, &t);
    f[a][b] = f[b][a] = min(f[a][b], t);
  }
  for (int k = 1; k <= v; ++k)
    for (int i = 1; i <= v; ++i)
      for (int j = 1; j <= v; ++j) f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
  int l = 0, r = 1731311, ans = -1;
  while (l <= r) {
    int mid = l + r >> 1;
    if (check(mid))
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%d", ans);
}

### 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;
namespace WXHAKIOI2018 {
struct edge {
  int r, nxt, w;
} e[700 * 600];
int S, T, head[1100], dep[1100], q[1100], l, r, esz, cur[1100];
void init(int s, int t) {
  esz = 1, S = s, T = t;
  for (int i = S; i <= T; ++i) head[i] = 0;
}
void addedge(int u, int v, int w) {
  e[++esz].r = v;
  e[esz].nxt = head[u];
  head[u] = esz;
  e[esz].w = w;
  e[++esz].r = u;
  e[esz].nxt = head[v];
  head[v] = esz;
  e[esz].w = 0;
}
int bfs() {
  for (int i = S; i <= T; ++i) dep[i] = 0, cur[i] = head[i];
  l = r = 0, dep[q[r++] = S] = 1;
  while (l < r) {
    int u = q[l++];
    for (int t = head[u]; t; t = e[t].nxt)
      if (e[t].w && !dep[e[t].r]) dep[q[r++] = e[t].r] = dep[u] + 1;
  }
  return dep[T] != 0;
}
int find(int u, int flow) {
  if (u == T) return flow;
  int used = 0, a = 0;
  for (int& t = cur[u]; t; t = e[t].nxt)
    if (e[t].w && dep[e[t].r] == dep[u] + 1) {
      a = find(e[t].r, min(flow - used, e[t].w));
      e[t].w -= a, e[t ^ 1].w += a, used += a;
      if (used == flow) return used;
    }
  if (!used) dep[u] = -1;
  return used;
}
int dinic() {
  int ans = 0, a;
  while (bfs())
    while (a = find(S, 1831311)) ans += a;
  return ans;
}
};  // namespace WXHAKIOI2018
int f[1100][1100], v, e, n, k;
int a[1100];
bool check(int ans) {
  WXHAKIOI2018::init(0, n + v + 1);
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= v; ++j)
      if (f[a[i]][j] <= ans) WXHAKIOI2018::addedge(i, j + n, 1831311);
  for (int i = 1; i <= n; ++i) WXHAKIOI2018::addedge(WXHAKIOI2018::S, i, 1);
  for (int i = 1; i <= v; ++i) WXHAKIOI2018::addedge(i + n, WXHAKIOI2018::T, 1);
  return WXHAKIOI2018::dinic() >= k;
}
int main() {
  scanf("%d%d%d%d", &v, &e, &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= v; ++i)
    for (int j = 1; j <= v; ++j)
      if (i != j) f[i][j] = 1831311;
  for (int i = 1, a, b, t; i <= e; ++i) {
    scanf("%d%d%d", &a, &b, &t);
    f[a][b] = f[b][a] = min(f[a][b], t);
  }
  for (int k = 1; k <= v; ++k)
    for (int i = 1; i <= v; ++i)
      for (int j = 1; j <= v; ++j) f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
  int l = 0, r = 1731311, ans = -1;
  while (l <= r) {
    int mid = l + r >> 1;
    if (check(mid))
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%d", ans);
}

```