#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 25;
struct Karzanov {
  struct edge {
    int t, c, r;
    edge(int _t, int _c, int _r) : t(_t), c(_c), r(_r) {}
  };
  vector<int> l, L;
  vector<vector<edge>> e;
  vector<int> ex;
  vector<vector<vector<int>>> st;
  Karzanov(int n) : e(n + 1), ex(n + 1), st(n + 1) {}
  void add(int u, int v, int w) {
    e[u].push_back(edge(v, w, e[v].size()));
    e[v].push_back(edge(u, 0, e[u].size() - 1));
  }
  edge& rev(const edge& E) { return e[E.t][E.r]; }
  bool bfs(int s, int t) {
    l.assign(e.size(), INF);
    l[s] = 1;
    queue<int> Q;
    for (Q.push(s); !Q.empty(); Q.pop()) {
      s = Q.front();
      L.push_back(s);
      for (const edge& E : e[s])
        if (E.c > 0 && l[E.t] > l[s] + 1) {
          l[E.t] = l[s] + 1;
          Q.push(E.t);
        }
    }
    return l[t] < INF;
  }
  void ch(int u, int p, int val) {
    int v = e[u][p].t, r = e[u][p].r;
    e[u][p].c -= val;
    e[v][r].c += val;
    ex[u] -= val;
    ex[v] += val;
  }
  void push(int u, int p, int val) {
    if (!val) return;
    int v = e[u][p].t, r = e[u][p].r;
    ch(u, p, val);
    st[v].push_back({u, p, val});
  }
  int operator()(int s, int t) {
    int ans = 0;
    for (; bfs(s, t);) {
      vector<bool> dead(l.size()), in(l.size());
      while (L.back() != t) L.pop_back();
      L.pop_back();
      reverse(L.begin(), L.end());
      L.pop_back();
      reverse(L.begin(), L.end());
      for (int u : L) in[u] = 1;
      in[s] = in[t] = 1;
      for (int i = 0; i < e[s].size(); i++) push(s, i, e[s][i].c);
      while (1) {
        for (int u : L) {
          if (dead[u] || ex[u] <= 0) continue;
          for (int i = 0; i < e[u].size() && ex[u] > 0; i++) {
            edge& E = e[u][i];
            if (E.c > 0 && l[u] + 1 == l[E.t] && !dead[E.t] && in[E.t]) {
              int val = min(ex[u], E.c);
              push(u, i, val);
              if (E.t == t) ans += val;
            }
          }
        }
        int u = -1;
        for (auto it = L.rbegin(); it != L.rend(); it++)
          if (!dead[*it] && ex[*it] > 0) {
            u = *it;
            break;
          }
        if (u == -1) break;
        while (ex[u] > 0) {
          int v = st[u].back()[0], p = st[u].back()[1];
          int val = min(ex[u], st[u].back()[2]);
          ch(v, p, -val);
          st[u].back()[2] -= val;
          if (st[u].back()[2] == 0) st[u].pop_back();
        }
        dead[u] = 1;
      }
      L.clear();
      ex.assign(l.size(), 0);
      st.assign(l.size(), vector<vector<int>>());
    }
    return ans;
  }
};
int main() {
  int n, m, x, y;
  scanf("%d%d%d%d", &n, &m, &x, &y);
  vector<vector<int>> e(n + 1, vector<int>(n + 1, INF));
  int s = 0, t = n + n + 1, ans = -1;
  Karzanov D(n + n + 2);
  for (; x > 0; x--) {
    int v;
    scanf("%d", &v);
    D.add(s, v, 1);
  }
  for (; m > 0; m--) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    e[u][v] = min(e[u][v], w);
    e[v][u] = min(e[v][u], w);
  }
  for (int i = 1; i <= n; i++) {
    e[i][i] = 0;
    D.add(n + i, t, 1);
  }
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) e[i][j] = min(e[i][j], e[i][k] + e[k][j]);
  for (int l = 0, r = INF; l < r;) {
    m = (l + r) >> 1;
    Karzanov P = D;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        if (e[i][j] <= m) P.add(i, n + j, 1);
    if (P(s, t) >= y)
      r = ans = m;
    else
      l = m + 1;
  }
  printf("%d\n", ans);
}

### Prompt

Your challenge is to write 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 = 1 << 25;
struct Karzanov {
  struct edge {
    int t, c, r;
    edge(int _t, int _c, int _r) : t(_t), c(_c), r(_r) {}
  };
  vector<int> l, L;
  vector<vector<edge>> e;
  vector<int> ex;
  vector<vector<vector<int>>> st;
  Karzanov(int n) : e(n + 1), ex(n + 1), st(n + 1) {}
  void add(int u, int v, int w) {
    e[u].push_back(edge(v, w, e[v].size()));
    e[v].push_back(edge(u, 0, e[u].size() - 1));
  }
  edge& rev(const edge& E) { return e[E.t][E.r]; }
  bool bfs(int s, int t) {
    l.assign(e.size(), INF);
    l[s] = 1;
    queue<int> Q;
    for (Q.push(s); !Q.empty(); Q.pop()) {
      s = Q.front();
      L.push_back(s);
      for (const edge& E : e[s])
        if (E.c > 0 && l[E.t] > l[s] + 1) {
          l[E.t] = l[s] + 1;
          Q.push(E.t);
        }
    }
    return l[t] < INF;
  }
  void ch(int u, int p, int val) {
    int v = e[u][p].t, r = e[u][p].r;
    e[u][p].c -= val;
    e[v][r].c += val;
    ex[u] -= val;
    ex[v] += val;
  }
  void push(int u, int p, int val) {
    if (!val) return;
    int v = e[u][p].t, r = e[u][p].r;
    ch(u, p, val);
    st[v].push_back({u, p, val});
  }
  int operator()(int s, int t) {
    int ans = 0;
    for (; bfs(s, t);) {
      vector<bool> dead(l.size()), in(l.size());
      while (L.back() != t) L.pop_back();
      L.pop_back();
      reverse(L.begin(), L.end());
      L.pop_back();
      reverse(L.begin(), L.end());
      for (int u : L) in[u] = 1;
      in[s] = in[t] = 1;
      for (int i = 0; i < e[s].size(); i++) push(s, i, e[s][i].c);
      while (1) {
        for (int u : L) {
          if (dead[u] || ex[u] <= 0) continue;
          for (int i = 0; i < e[u].size() && ex[u] > 0; i++) {
            edge& E = e[u][i];
            if (E.c > 0 && l[u] + 1 == l[E.t] && !dead[E.t] && in[E.t]) {
              int val = min(ex[u], E.c);
              push(u, i, val);
              if (E.t == t) ans += val;
            }
          }
        }
        int u = -1;
        for (auto it = L.rbegin(); it != L.rend(); it++)
          if (!dead[*it] && ex[*it] > 0) {
            u = *it;
            break;
          }
        if (u == -1) break;
        while (ex[u] > 0) {
          int v = st[u].back()[0], p = st[u].back()[1];
          int val = min(ex[u], st[u].back()[2]);
          ch(v, p, -val);
          st[u].back()[2] -= val;
          if (st[u].back()[2] == 0) st[u].pop_back();
        }
        dead[u] = 1;
      }
      L.clear();
      ex.assign(l.size(), 0);
      st.assign(l.size(), vector<vector<int>>());
    }
    return ans;
  }
};
int main() {
  int n, m, x, y;
  scanf("%d%d%d%d", &n, &m, &x, &y);
  vector<vector<int>> e(n + 1, vector<int>(n + 1, INF));
  int s = 0, t = n + n + 1, ans = -1;
  Karzanov D(n + n + 2);
  for (; x > 0; x--) {
    int v;
    scanf("%d", &v);
    D.add(s, v, 1);
  }
  for (; m > 0; m--) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    e[u][v] = min(e[u][v], w);
    e[v][u] = min(e[v][u], w);
  }
  for (int i = 1; i <= n; i++) {
    e[i][i] = 0;
    D.add(n + i, t, 1);
  }
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) e[i][j] = min(e[i][j], e[i][k] + e[k][j]);
  for (int l = 0, r = INF; l < r;) {
    m = (l + r) >> 1;
    Karzanov P = D;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        if (e[i][j] <= m) P.add(i, n + j, 1);
    if (P(s, t) >= y)
      r = ans = m;
    else
      l = m + 1;
  }
  printf("%d\n", ans);
}

```