#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int N = 1505, M = 1000005;
const int TEAM = 700;
int G[N][N], pos[N];
int n, m, k, limit;
int s = N - 1, t = N - 2, tot;
int head[N], dis[N], cur[N];
struct node {
  int v, f;
  int next;
} e[M];
void init() {
  tot = 0;
  memset(head, -1, sizeof head);
}
void add(int u, int v, int f) {
  e[tot].v = v;
  e[tot].f = f;
  e[tot].next = head[u];
  head[u] = tot++;
}
void addedge(int u, int v, int f) {
  add(u, v, f);
  add(v, u, 0);
}
int bfs(int ds, int t) {
  memset(dis, -1, sizeof(dis));
  dis[ds] = 0;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (u == t) return 1;
    for (int i = head[u]; ~i; i = e[i].next) {
      int v = e[i].v;
      if (dis[v] == -1 && e[i].f) {
        dis[v] = dis[u] + 1;
        q.push(v);
        if (v == t) break;
      }
    }
  }
  return dis[t] > 0;
}
int dfs(int ds, int t, int min1) {
  if (ds == t) return min1;
  int flow, ans = 0;
  for (int i = cur[ds]; ~i; i = e[i].next) {
    int v = e[i].v;
    cur[ds] = i;
    if (dis[v] == dis[ds] + 1 && e[i].f) {
      if (flow = dfs(v, t, min(min1 - ans, e[i].f))) {
        e[i].f -= flow;
        e[i ^ 1].f += flow;
        ans += flow;
        if (min1 == ans) break;
      }
    }
  }
  if (!ans) dis[ds] = -1;
  return ans;
}
int dinic() {
  int ans = 0, x;
  while (bfs(s, t)) {
    memcpy(cur, head, sizeof(head));
    while (x = dfs(s, t, inf)) ans += x;
  }
  return ans;
}
bool pd(int x) {
  init();
  for (int i = 1; i <= k; ++i) {
    addedge(s, i + TEAM, 1);
    for (int j = 1; j <= n; ++j) {
      if (G[pos[i]][j] <= x) addedge(i + TEAM, j, 1);
    }
  }
  for (int i = 1; i <= n; ++i) addedge(i, t, 1);
  if (dinic() >= limit)
    return true;
  else
    return false;
}
int main() {
  int x, y, z;
  scanf("%d%d%d%d", &n, &m, &k, &limit);
  for (int i = 1; i <= k; ++i) scanf("%d", &pos[i]);
  memset(G, inf, sizeof(G));
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d", &x, &y, &z);
    G[x][y] = G[y][x] = min(z, G[x][y]);
  }
  if (limit > k) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; ++i) G[i][i] = 0;
  for (int k = 1; k <= n; ++k)
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= n; ++j) G[i][j] = min(G[i][j], G[i][k] + G[k][j]);
  int l = 0, r = 1731311, mid;
  int ans = -1;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (pd(mid))
      r = mid - 1, ans = mid;
    else
      l = mid + 1;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

In CPP, your task is to solve 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 N = 1505, M = 1000005;
const int TEAM = 700;
int G[N][N], pos[N];
int n, m, k, limit;
int s = N - 1, t = N - 2, tot;
int head[N], dis[N], cur[N];
struct node {
  int v, f;
  int next;
} e[M];
void init() {
  tot = 0;
  memset(head, -1, sizeof head);
}
void add(int u, int v, int f) {
  e[tot].v = v;
  e[tot].f = f;
  e[tot].next = head[u];
  head[u] = tot++;
}
void addedge(int u, int v, int f) {
  add(u, v, f);
  add(v, u, 0);
}
int bfs(int ds, int t) {
  memset(dis, -1, sizeof(dis));
  dis[ds] = 0;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (u == t) return 1;
    for (int i = head[u]; ~i; i = e[i].next) {
      int v = e[i].v;
      if (dis[v] == -1 && e[i].f) {
        dis[v] = dis[u] + 1;
        q.push(v);
        if (v == t) break;
      }
    }
  }
  return dis[t] > 0;
}
int dfs(int ds, int t, int min1) {
  if (ds == t) return min1;
  int flow, ans = 0;
  for (int i = cur[ds]; ~i; i = e[i].next) {
    int v = e[i].v;
    cur[ds] = i;
    if (dis[v] == dis[ds] + 1 && e[i].f) {
      if (flow = dfs(v, t, min(min1 - ans, e[i].f))) {
        e[i].f -= flow;
        e[i ^ 1].f += flow;
        ans += flow;
        if (min1 == ans) break;
      }
    }
  }
  if (!ans) dis[ds] = -1;
  return ans;
}
int dinic() {
  int ans = 0, x;
  while (bfs(s, t)) {
    memcpy(cur, head, sizeof(head));
    while (x = dfs(s, t, inf)) ans += x;
  }
  return ans;
}
bool pd(int x) {
  init();
  for (int i = 1; i <= k; ++i) {
    addedge(s, i + TEAM, 1);
    for (int j = 1; j <= n; ++j) {
      if (G[pos[i]][j] <= x) addedge(i + TEAM, j, 1);
    }
  }
  for (int i = 1; i <= n; ++i) addedge(i, t, 1);
  if (dinic() >= limit)
    return true;
  else
    return false;
}
int main() {
  int x, y, z;
  scanf("%d%d%d%d", &n, &m, &k, &limit);
  for (int i = 1; i <= k; ++i) scanf("%d", &pos[i]);
  memset(G, inf, sizeof(G));
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d", &x, &y, &z);
    G[x][y] = G[y][x] = min(z, G[x][y]);
  }
  if (limit > k) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; ++i) G[i][i] = 0;
  for (int k = 1; k <= n; ++k)
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= n; ++j) G[i][j] = min(G[i][j], G[i][k] + G[k][j]);
  int l = 0, r = 1731311, mid;
  int ans = -1;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (pd(mid))
      r = mid - 1, ans = mid;
    else
      l = mid + 1;
  }
  printf("%d\n", ans);
  return 0;
}

```