#include <bits/stdc++.h>
using namespace std;
char buf[1 << 21], *p1 = buf, *p2 = buf;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = (p1 == p2 &&
                    (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
                ? EOF
                : *p1++),
       f = 0;
  for (; c < '0' || c > '9';
       c = (p1 == p2 &&
                    (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
                ? EOF
                : *p1++))
    if (!(c ^ 45)) f = 1;
  for (; c >= '0' && c <= '9';
       c = (p1 == p2 &&
                    (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
                ? EOF
                : *p1++))
    x = (x << 1) + (x << 3) + (c ^ 48);
  f ? x = -x, 1 : 1;
}
const int INF = 1e9 + 5;
struct edge {
  int to;
  int w;
  int nxt;
} e[1540005];
int s, t, et, head[1205], cur[1205], d[1205];
inline void ADDE(int x, int y, int w) {
  e[++et] = (edge){y, w, head[x]}, head[x] = et;
}
inline void adde(int x, int y, int w) { ADDE(x, y, w), ADDE(y, x, 0); }
inline char bfs(int s, int t) {
  queue<int> q;
  q.push(s), memset(d, 0, sizeof(d)), d[s] = 1;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; i; i = e[i].nxt)
      if (!d[e[i].to] && e[i].w) q.push(e[i].to), d[e[i].to] = d[x] + 1;
  }
  return !!d[t];
}
inline int dfs(int x, int t, int lim = INF) {
  int f = lim;
  if (x == t) return lim;
  for (int &i = cur[x]; i; i = e[i].nxt)
    if (d[e[i].to] == d[x] + 1 && e[i].w) {
      int g = dfs(e[i].to, t, min(f, e[i].w));
      e[i].w -= g, e[((((i)&1) ? 1 : -1) + (i))].w += g, f -= g;
      if (!f) break;
    }
  return lim - f;
}
inline int dinic(int s, int t) {
  int r = 0;
  while (bfs(s, t)) memcpy(cur, head, sizeof(cur)), r += dfs(s, t);
  return r;
}
int n, m, ct, K, w[605], ds[605][605];
inline char check(int md) {
  et = 0, memset(head, 0, sizeof(head));
  for (int i = 1; i <= n; i++)
    if (w[i]) adde(s, i, w[i]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (ds[i][j] <= md) adde(i, j + n, INF);
  for (int i = 1; i <= n; i++) adde(i + n, t, 1);
  return dinic(s, t) >= K;
}
int main() {
  read(n), read(m), read(ct), read(K), s = n << 1 | 1, t = s + 1,
                                       memset(ds, 0x3f, sizeof(ds));
  for (int i = 1; i <= n; i++) ds[i][i] = 0;
  for (int i = 1, x; i <= ct; i++) read(x), w[x]++;
  for (int i = 1, x, y, w; i <= m; i++)
    read(x), read(y), read(w), ds[x][y] = ds[y][x] = min(ds[y][x], w);
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        ds[i][j] = min(ds[i][j], ds[i][k] + ds[k][j]);
  int le = 0, ri = 1731311, md, rs = ri + 1;
  while (le <= ri) {
    md = (le + ri) >> 1;
    if (check(md))
      rs = md, ri = md - 1;
    else
      le = md + 1;
  }
  return printf("%d\n", rs > 1731311 ? -1 : rs), 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
char buf[1 << 21], *p1 = buf, *p2 = buf;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = (p1 == p2 &&
                    (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
                ? EOF
                : *p1++),
       f = 0;
  for (; c < '0' || c > '9';
       c = (p1 == p2 &&
                    (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
                ? EOF
                : *p1++))
    if (!(c ^ 45)) f = 1;
  for (; c >= '0' && c <= '9';
       c = (p1 == p2 &&
                    (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
                ? EOF
                : *p1++))
    x = (x << 1) + (x << 3) + (c ^ 48);
  f ? x = -x, 1 : 1;
}
const int INF = 1e9 + 5;
struct edge {
  int to;
  int w;
  int nxt;
} e[1540005];
int s, t, et, head[1205], cur[1205], d[1205];
inline void ADDE(int x, int y, int w) {
  e[++et] = (edge){y, w, head[x]}, head[x] = et;
}
inline void adde(int x, int y, int w) { ADDE(x, y, w), ADDE(y, x, 0); }
inline char bfs(int s, int t) {
  queue<int> q;
  q.push(s), memset(d, 0, sizeof(d)), d[s] = 1;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; i; i = e[i].nxt)
      if (!d[e[i].to] && e[i].w) q.push(e[i].to), d[e[i].to] = d[x] + 1;
  }
  return !!d[t];
}
inline int dfs(int x, int t, int lim = INF) {
  int f = lim;
  if (x == t) return lim;
  for (int &i = cur[x]; i; i = e[i].nxt)
    if (d[e[i].to] == d[x] + 1 && e[i].w) {
      int g = dfs(e[i].to, t, min(f, e[i].w));
      e[i].w -= g, e[((((i)&1) ? 1 : -1) + (i))].w += g, f -= g;
      if (!f) break;
    }
  return lim - f;
}
inline int dinic(int s, int t) {
  int r = 0;
  while (bfs(s, t)) memcpy(cur, head, sizeof(cur)), r += dfs(s, t);
  return r;
}
int n, m, ct, K, w[605], ds[605][605];
inline char check(int md) {
  et = 0, memset(head, 0, sizeof(head));
  for (int i = 1; i <= n; i++)
    if (w[i]) adde(s, i, w[i]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (ds[i][j] <= md) adde(i, j + n, INF);
  for (int i = 1; i <= n; i++) adde(i + n, t, 1);
  return dinic(s, t) >= K;
}
int main() {
  read(n), read(m), read(ct), read(K), s = n << 1 | 1, t = s + 1,
                                       memset(ds, 0x3f, sizeof(ds));
  for (int i = 1; i <= n; i++) ds[i][i] = 0;
  for (int i = 1, x; i <= ct; i++) read(x), w[x]++;
  for (int i = 1, x, y, w; i <= m; i++)
    read(x), read(y), read(w), ds[x][y] = ds[y][x] = min(ds[y][x], w);
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        ds[i][j] = min(ds[i][j], ds[i][k] + ds[k][j]);
  int le = 0, ri = 1731311, md, rs = ri + 1;
  while (le <= ri) {
    md = (le + ri) >> 1;
    if (check(md))
      rs = md, ri = md - 1;
    else
      le = md + 1;
  }
  return printf("%d\n", rs > 1731311 ? -1 : rs), 0;
}

```