#include <bits/stdc++.h>
using namespace std;
inline int gi() {
  int f = 1, x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f * x;
}
const int maxn = 20003;
int n, m, x, w, v, k, tot, head[maxn], ver[maxn * 2], nxt[maxn * 2],
    edge[maxn * 2];
int e, a[maxn], dp[603][603], g[603][603], vis[maxn], match[maxn];
inline void add(int u, int v, int w) {
  ver[++tot] = v, edge[tot] = w, nxt[tot] = head[u], head[u] = tot;
}
bool dfs(int u) {
  for (int i = 1; i <= v; i += 1) {
    if (g[u][i] && !vis[i]) {
      vis[i] = 1;
      if (!match[i] || dfs(match[i])) {
        match[i] = u;
        return true;
      }
    }
  }
  return false;
}
inline int pp() {
  int uu = 0;
  memset(match, 0, sizeof(match));
  for (int i = 1; i <= n; i += 1) {
    memset(vis, 0, sizeof(vis));
    if (dfs(i)) ++uu;
  }
  return uu;
}
inline bool check(int mid) {
  memset(g, 0, sizeof(g));
  for (int i = 1; i <= n; i += 1) {
    for (int j = 1; j <= v; j += 1) {
      if (dp[a[i]][j] <= mid) g[i][j] = 1;
    }
  }
  int ans = pp();
  return ans >= k;
}
int main() {
  v = gi(), e = gi(), n = gi(), k = gi();
  memset(dp, 0x3f, sizeof(dp));
  for (int i = 1; i <= n; i += 1) {
    a[i] = gi();
  }
  for (int i = 1; i <= v; i += 1) dp[i][i] = 0;
  for (int i = 1; i <= e; i += 1) {
    int u = gi(), vv = gi(), w = gi();
    add(u, vv, w), add(vv, u, w);
    dp[u][vv] = dp[vv][u] = min(dp[u][vv], w);
  }
  for (int k = 1; k <= v; k += 1) {
    for (int i = 1; i <= v; i += 1) {
      for (int j = 1; j <= v; j += 1) {
        dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);
      }
    }
  }
  int l = 0, r = 1731319, su = -1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      su = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%d\n", su);
  return 0;
}

### Prompt

Please formulate 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;
inline int gi() {
  int f = 1, x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f * x;
}
const int maxn = 20003;
int n, m, x, w, v, k, tot, head[maxn], ver[maxn * 2], nxt[maxn * 2],
    edge[maxn * 2];
int e, a[maxn], dp[603][603], g[603][603], vis[maxn], match[maxn];
inline void add(int u, int v, int w) {
  ver[++tot] = v, edge[tot] = w, nxt[tot] = head[u], head[u] = tot;
}
bool dfs(int u) {
  for (int i = 1; i <= v; i += 1) {
    if (g[u][i] && !vis[i]) {
      vis[i] = 1;
      if (!match[i] || dfs(match[i])) {
        match[i] = u;
        return true;
      }
    }
  }
  return false;
}
inline int pp() {
  int uu = 0;
  memset(match, 0, sizeof(match));
  for (int i = 1; i <= n; i += 1) {
    memset(vis, 0, sizeof(vis));
    if (dfs(i)) ++uu;
  }
  return uu;
}
inline bool check(int mid) {
  memset(g, 0, sizeof(g));
  for (int i = 1; i <= n; i += 1) {
    for (int j = 1; j <= v; j += 1) {
      if (dp[a[i]][j] <= mid) g[i][j] = 1;
    }
  }
  int ans = pp();
  return ans >= k;
}
int main() {
  v = gi(), e = gi(), n = gi(), k = gi();
  memset(dp, 0x3f, sizeof(dp));
  for (int i = 1; i <= n; i += 1) {
    a[i] = gi();
  }
  for (int i = 1; i <= v; i += 1) dp[i][i] = 0;
  for (int i = 1; i <= e; i += 1) {
    int u = gi(), vv = gi(), w = gi();
    add(u, vv, w), add(vv, u, w);
    dp[u][vv] = dp[vv][u] = min(dp[u][vv], w);
  }
  for (int k = 1; k <= v; k += 1) {
    for (int i = 1; i <= v; i += 1) {
      for (int j = 1; j <= v; j += 1) {
        dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);
      }
    }
  }
  int l = 0, r = 1731319, su = -1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      su = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%d\n", su);
  return 0;
}

```