#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1205;
const long long INF = 1000000000000ll;
long long n, c[MAXN][MAXN], f[MAXN][MAXN], s, t, d[MAXN], ptr[MAXN], q[MAXN];
long long dr[605][605];
bool bfs() {
  long long qh = 0, qt = 0;
  q[qt++] = s;
  memset(d, -1, n * sizeof d[0]);
  d[s] = 0;
  while (qh < qt) {
    long long v = q[qh++];
    for (long long to = 0; to < n; ++to)
      if (d[to] == -1 && f[v][to] < c[v][to]) {
        q[qt++] = to;
        d[to] = d[v] + 1;
      }
  }
  return d[t] != -1;
}
long long dfs(long long v, long long flow) {
  if (!flow) return 0;
  if (v == t) return flow;
  for (long long& to = ptr[v]; to < n; ++to) {
    if (d[to] != d[v] + 1) continue;
    long long pushed = dfs(to, min(flow, c[v][to] - f[v][to]));
    if (pushed) {
      f[v][to] += pushed;
      f[to][v] -= pushed;
      return pushed;
    }
  }
  return 0;
}
long long dinic() {
  long long flow = 0;
  for (;;) {
    if (!bfs()) break;
    memset(ptr, 0, n * sizeof ptr[0]);
    while (long long pushed = dfs(s, INF)) flow += pushed;
  }
  return flow;
}
int main() {
  {
    long long v, e, nn, k;
    cin >> v >> e >> nn >> k;
    s = 0, t = 1 + 2 * v, n = 2 + 2 * v;
    vector<long long> vr;
    for (int i = 1; i < nn + 1; i++) {
      long long x;
      cin >> x;
      vr.push_back(x);
    }
    for (int i = 1; i < v + 1; i++)
      for (int j = 1; j < v + 1; j++) dr[i][j] = INF;
    for (int i = 0; i < e; i++) {
      long long x, y, z;
      cin >> x >> y >> z;
      dr[x][y] = min(dr[x][y], z);
      dr[y][x] = min(dr[y][x], z);
    }
    for (int i = 1; i < v + 1; i++) dr[i][i] = 0;
    for (int k = 1; k < v + 1; k++) {
      for (int i = 1; i < v + 1; i++) {
        for (int j = 1; j < v + 1; j++) {
          if (dr[i][k] + dr[k][j] < dr[i][j]) dr[i][j] = dr[i][k] + dr[k][j];
        }
      }
    }
    long long low = 0, high = 1731311, mid = 3, ans = -1;
    for (int i = 0; i < vr.size(); i++) {
      c[0][vr[i]]++;
    }
    for (int i = 1; i < v + 1; i++) {
      c[v + i][2 * v + 1] = 1;
    }
    while (low <= high) {
      mid = (low + high) / 2;
      for (int i = 1; i < v + 1; i++) {
        for (int j = 1; j < v + 1; j++) {
          if (dr[i][j] <= mid)
            c[i][v + j] = 1;
          else
            c[i][v + j] = 0;
        }
      }
      if (dinic() >= k) {
        ans = mid;
        high = mid - 1;
      } else
        low = mid + 1;
      memset(f, 0, sizeof(f));
      memset(d, 0, sizeof(d));
      memset(q, 0, sizeof(q));
      memset(ptr, 0, sizeof(ptr));
    }
    cout << ans << " ";
  }
}

### 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 long long MAXN = 1205;
const long long INF = 1000000000000ll;
long long n, c[MAXN][MAXN], f[MAXN][MAXN], s, t, d[MAXN], ptr[MAXN], q[MAXN];
long long dr[605][605];
bool bfs() {
  long long qh = 0, qt = 0;
  q[qt++] = s;
  memset(d, -1, n * sizeof d[0]);
  d[s] = 0;
  while (qh < qt) {
    long long v = q[qh++];
    for (long long to = 0; to < n; ++to)
      if (d[to] == -1 && f[v][to] < c[v][to]) {
        q[qt++] = to;
        d[to] = d[v] + 1;
      }
  }
  return d[t] != -1;
}
long long dfs(long long v, long long flow) {
  if (!flow) return 0;
  if (v == t) return flow;
  for (long long& to = ptr[v]; to < n; ++to) {
    if (d[to] != d[v] + 1) continue;
    long long pushed = dfs(to, min(flow, c[v][to] - f[v][to]));
    if (pushed) {
      f[v][to] += pushed;
      f[to][v] -= pushed;
      return pushed;
    }
  }
  return 0;
}
long long dinic() {
  long long flow = 0;
  for (;;) {
    if (!bfs()) break;
    memset(ptr, 0, n * sizeof ptr[0]);
    while (long long pushed = dfs(s, INF)) flow += pushed;
  }
  return flow;
}
int main() {
  {
    long long v, e, nn, k;
    cin >> v >> e >> nn >> k;
    s = 0, t = 1 + 2 * v, n = 2 + 2 * v;
    vector<long long> vr;
    for (int i = 1; i < nn + 1; i++) {
      long long x;
      cin >> x;
      vr.push_back(x);
    }
    for (int i = 1; i < v + 1; i++)
      for (int j = 1; j < v + 1; j++) dr[i][j] = INF;
    for (int i = 0; i < e; i++) {
      long long x, y, z;
      cin >> x >> y >> z;
      dr[x][y] = min(dr[x][y], z);
      dr[y][x] = min(dr[y][x], z);
    }
    for (int i = 1; i < v + 1; i++) dr[i][i] = 0;
    for (int k = 1; k < v + 1; k++) {
      for (int i = 1; i < v + 1; i++) {
        for (int j = 1; j < v + 1; j++) {
          if (dr[i][k] + dr[k][j] < dr[i][j]) dr[i][j] = dr[i][k] + dr[k][j];
        }
      }
    }
    long long low = 0, high = 1731311, mid = 3, ans = -1;
    for (int i = 0; i < vr.size(); i++) {
      c[0][vr[i]]++;
    }
    for (int i = 1; i < v + 1; i++) {
      c[v + i][2 * v + 1] = 1;
    }
    while (low <= high) {
      mid = (low + high) / 2;
      for (int i = 1; i < v + 1; i++) {
        for (int j = 1; j < v + 1; j++) {
          if (dr[i][j] <= mid)
            c[i][v + j] = 1;
          else
            c[i][v + j] = 0;
        }
      }
      if (dinic() >= k) {
        ans = mid;
        high = mid - 1;
      } else
        low = mid + 1;
      memset(f, 0, sizeof(f));
      memset(d, 0, sizeof(d));
      memset(q, 0, sizeof(q));
      memset(ptr, 0, sizeof(ptr));
    }
    cout << ans << " ";
  }
}

```