#include <bits/stdc++.h>
using namespace std;
uint64_t rnd_data = 0xDEADBEEFDULL;
inline void my_srand(int seed) { rnd_data = ((uint64_t)seed << 16) | 0x330E; }
inline int my_rand() {
  rnd_data = rnd_data * 0x5DEECE66DULL + 0xB;
  return (rnd_data >> 17) & 0x7FFFFFFF;
}
template <typename T>
void my_random_shuffle(T b, T e) {
  for (ssize_t i = 1; i < (e - b); i++) {
    swap(b[i], b[my_rand() % (i + 1)]);
  }
}
template <class _T>
inline _T sqr(const _T &x) {
  return x * x;
}
const long double PI = 3.1415926535897932384626433832795;
const long double EPS = 1e-9;
const int MAXN = 1024;
int n, m, k, t;
bool a[MAXN][MAXN];
int d[MAXN][MAXN];
int c[MAXN];
bool u[MAXN];
int pa[MAXN];
bool find(int k) {
  u[k] = true;
  for (int i = 0; i < (int)(n); i++) {
    if (a[k][i] && (pa[i] == -1 || (!u[pa[i]] && find(pa[i])))) {
      pa[i] = k;
      return true;
    }
  }
  return false;
}
int calc() {
  memset(pa, 0xff, sizeof(pa));
  int ans = 0;
  memset((u), 0, sizeof(u));
  for (int i = 0; i < (int)(n); i++) {
    for (int j = 0; j < (int)(c[i]); j++) {
      if (find(i)) {
        ans++;
        memset((u), 0, sizeof(u));
      } else
        break;
    }
  }
  return ans;
}
int main() {
  cout << setiosflags(ios::fixed) << setprecision(10);
  memset(d, 0x3f, sizeof(d));
  memset((c), 0, sizeof(c));
  scanf("%d%d%d%d", &n, &m, &t, &k);
  for (int i = 0; i < (int)(t); i++) {
    int x;
    scanf("%d", &x);
    x--;
    c[x]++;
  }
  for (int i = 0; i < (int)(n); i++) {
    d[i][i] = 0;
  }
  for (int i = 0; i < (int)(m); i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    x--;
    y--;
    d[x][y] = d[y][x] = min(d[x][y], z);
  }
  for (int k = 0; k < (int)(n); k++) {
    for (int i = 0; i < (int)(n); i++) {
      for (int j = 0; j < (int)(n); j++) {
        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
      }
    }
  }
  int mi = 0;
  int ma = 1731312;
  while (mi < ma) {
    int q = (mi + ma) / 2;
    for (int i = 0; i < (int)(n); i++) {
      for (int j = 0; j < (int)(n); j++) {
        a[i][j] = d[i][j] <= q;
      }
    }
    if (calc() >= k)
      ma = q;
    else
      mi = q + 1;
  }
  if (mi >= 1731312)
    puts("-1");
  else
    printf("%d\n", mi);
  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;
uint64_t rnd_data = 0xDEADBEEFDULL;
inline void my_srand(int seed) { rnd_data = ((uint64_t)seed << 16) | 0x330E; }
inline int my_rand() {
  rnd_data = rnd_data * 0x5DEECE66DULL + 0xB;
  return (rnd_data >> 17) & 0x7FFFFFFF;
}
template <typename T>
void my_random_shuffle(T b, T e) {
  for (ssize_t i = 1; i < (e - b); i++) {
    swap(b[i], b[my_rand() % (i + 1)]);
  }
}
template <class _T>
inline _T sqr(const _T &x) {
  return x * x;
}
const long double PI = 3.1415926535897932384626433832795;
const long double EPS = 1e-9;
const int MAXN = 1024;
int n, m, k, t;
bool a[MAXN][MAXN];
int d[MAXN][MAXN];
int c[MAXN];
bool u[MAXN];
int pa[MAXN];
bool find(int k) {
  u[k] = true;
  for (int i = 0; i < (int)(n); i++) {
    if (a[k][i] && (pa[i] == -1 || (!u[pa[i]] && find(pa[i])))) {
      pa[i] = k;
      return true;
    }
  }
  return false;
}
int calc() {
  memset(pa, 0xff, sizeof(pa));
  int ans = 0;
  memset((u), 0, sizeof(u));
  for (int i = 0; i < (int)(n); i++) {
    for (int j = 0; j < (int)(c[i]); j++) {
      if (find(i)) {
        ans++;
        memset((u), 0, sizeof(u));
      } else
        break;
    }
  }
  return ans;
}
int main() {
  cout << setiosflags(ios::fixed) << setprecision(10);
  memset(d, 0x3f, sizeof(d));
  memset((c), 0, sizeof(c));
  scanf("%d%d%d%d", &n, &m, &t, &k);
  for (int i = 0; i < (int)(t); i++) {
    int x;
    scanf("%d", &x);
    x--;
    c[x]++;
  }
  for (int i = 0; i < (int)(n); i++) {
    d[i][i] = 0;
  }
  for (int i = 0; i < (int)(m); i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    x--;
    y--;
    d[x][y] = d[y][x] = min(d[x][y], z);
  }
  for (int k = 0; k < (int)(n); k++) {
    for (int i = 0; i < (int)(n); i++) {
      for (int j = 0; j < (int)(n); j++) {
        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
      }
    }
  }
  int mi = 0;
  int ma = 1731312;
  while (mi < ma) {
    int q = (mi + ma) / 2;
    for (int i = 0; i < (int)(n); i++) {
      for (int j = 0; j < (int)(n); j++) {
        a[i][j] = d[i][j] <= q;
      }
    }
    if (calc() >= k)
      ma = q;
    else
      mi = q + 1;
  }
  if (mi >= 1731312)
    puts("-1");
  else
    printf("%d\n", mi);
  return 0;
}

```