#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long powmod(long long a, long long b) {
  long long ans = 1, pow = a % mod;
  while (b) {
    if (b & 1) ans = (ans * pow) % mod;
    pow = (pow * pow) % mod;
    b >>= 1;
  }
  return ans;
}
int v, e, n, k, l, r, mid, ans = -1;
int mp[1500][1500];
int s[1500];
int girl[1500];
int line[1500][1500];
bool used[1500];
void floyd() {
  for (int k = 1; k <= v; k++) {
    for (int i = 1; i <= v; i++) {
      for (int j = 1; j <= v; j++) {
        if (mp[i][j] > mp[i][k] + mp[k][j]) {
          mp[i][j] = mp[i][k] + mp[k][j];
        }
      }
    }
  }
}
bool match(int x) {
  for (int i = 1; i <= v; ++i) {
    if (line[x][i] && !used[i]) {
      used[i] = true;
      if (girl[i] == 0 || match(girl[i])) {
        girl[i] = x;
        return true;
      }
    }
  }
  return false;
}
int sssbbb() {
  memset(girl, 0, sizeof(girl));
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    memset(used, 0, sizeof(used));
    if (match(i)) ans++;
  }
  return ans;
}
bool judge(int x) {
  memset(line, 0, sizeof(line));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= v; j++) {
      if (mp[s[i]][j] <= x) line[i][j] = 1;
    }
  }
  int t = sssbbb();
  return (t >= k);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  memset(mp, inf, sizeof(mp));
  cin >> v >> e >> n >> k;
  for (int i = 1; i <= v; i++) mp[i][i] = 0;
  for (int i = 1; i <= n; ++i) cin >> s[i];
  for (int i = 1; i <= e; i++) {
    int x, y, w;
    cin >> x >> y >> w;
    mp[x][y] = mp[y][x] = min(mp[x][y], w);
  }
  floyd();
  l = 0, r = 1731311;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (judge(mid)) {
      ans = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  cout << ans << '\n';
}

### 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;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long powmod(long long a, long long b) {
  long long ans = 1, pow = a % mod;
  while (b) {
    if (b & 1) ans = (ans * pow) % mod;
    pow = (pow * pow) % mod;
    b >>= 1;
  }
  return ans;
}
int v, e, n, k, l, r, mid, ans = -1;
int mp[1500][1500];
int s[1500];
int girl[1500];
int line[1500][1500];
bool used[1500];
void floyd() {
  for (int k = 1; k <= v; k++) {
    for (int i = 1; i <= v; i++) {
      for (int j = 1; j <= v; j++) {
        if (mp[i][j] > mp[i][k] + mp[k][j]) {
          mp[i][j] = mp[i][k] + mp[k][j];
        }
      }
    }
  }
}
bool match(int x) {
  for (int i = 1; i <= v; ++i) {
    if (line[x][i] && !used[i]) {
      used[i] = true;
      if (girl[i] == 0 || match(girl[i])) {
        girl[i] = x;
        return true;
      }
    }
  }
  return false;
}
int sssbbb() {
  memset(girl, 0, sizeof(girl));
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    memset(used, 0, sizeof(used));
    if (match(i)) ans++;
  }
  return ans;
}
bool judge(int x) {
  memset(line, 0, sizeof(line));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= v; j++) {
      if (mp[s[i]][j] <= x) line[i][j] = 1;
    }
  }
  int t = sssbbb();
  return (t >= k);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  memset(mp, inf, sizeof(mp));
  cin >> v >> e >> n >> k;
  for (int i = 1; i <= v; i++) mp[i][i] = 0;
  for (int i = 1; i <= n; ++i) cin >> s[i];
  for (int i = 1; i <= e; i++) {
    int x, y, w;
    cin >> x >> y >> w;
    mp[x][y] = mp[y][x] = min(mp[x][y], w);
  }
  floyd();
  l = 0, r = 1731311;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (judge(mid)) {
      ans = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  cout << ans << '\n';
}

```