Masha wants to open her own bakery and bake muffins in one of the n cities numbered from 1 to n. There are m bidirectional roads, each of whose connects some pair of cities.

To bake muffins in her bakery, Masha needs to establish flour supply from some storage. There are only k storages, located in different cities numbered a1, a2, ..., ak.

Unforunately the law of the country Masha lives in prohibits opening bakery in any of the cities which has storage located in it. She can open it only in one of another n - k cities, and, of course, flour delivery should be paid — for every kilometer of path between storage and bakery Masha should pay 1 ruble.

Formally, Masha will pay x roubles, if she will open the bakery in some city b (ai ≠ b for every 1 ≤ i ≤ k) and choose a storage in some city s (s = aj for some 1 ≤ j ≤ k) and b and s are connected by some path of roads of summary length x (if there are more than one path, Masha is able to choose which of them should be used).

Masha is very thrifty and rational. She is interested in a city, where she can open her bakery (and choose one of k storages and one of the paths between city with bakery and city with storage) and pay minimum possible amount of rubles for flour delivery. Please help Masha find this amount.

Input

The first line of the input contains three integers n, m and k (1 ≤ n, m ≤ 105, 0 ≤ k ≤ n) — the number of cities in country Masha lives in, the number of roads between them and the number of flour storages respectively.

Then m lines follow. Each of them contains three integers u, v and l (1 ≤ u, v ≤ n, 1 ≤ l ≤ 109, u ≠ v) meaning that there is a road between cities u and v of length of l kilometers .

If k > 0, then the last line of the input contains k distinct integers a1, a2, ..., ak (1 ≤ ai ≤ n) — the number of cities having flour storage located in. If k = 0 then this line is not presented in the input.

Output

Print the minimum possible amount of rubles Masha should pay for flour delivery in the only line.

If the bakery can not be opened (while satisfying conditions) in any of the n cities, print  - 1 in the only line.

Examples

Input

5 4 2
1 2 5
1 2 3
2 3 4
1 4 10
1 5


Output

3

Input

3 1 1
1 2 3
3


Output

-1

Note

<image>

Image illustrates the first sample case. Cities with storage located in and the road representing the answer are darkened. 
import java.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Random;

public class ProblemB {
  
  public static void main(String[] args) throws IOException {
    init();
    new ProblemB().run();
    out.flush();
    out.close();
  }
  
  static void init() throws IOException {
    in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    out = new PrintWriter(new OutputStreamWriter(System.out));
    in.ordinaryChars(0, 65535);
    in.wordChars(0, 65535);
    in.whitespaceChars(' ', ' ');
    in.whitespaceChars('\n', '\n');
    in.whitespaceChars('\r', '\r');
  }
  
  // класс для пар
  class Pair<L, R> {
    private final L X;
    private final R Y;
    
    public Pair(L X, R Y) {
      this.X = X;
      this.Y = Y;
    }
    
    public L getX() {
      return X;
    }
    
    public R getY() {
      return Y;
    }
    
    @Override
    public int hashCode() {
      return X.hashCode() ^ Y.hashCode();
    }
    
    @Override
    public boolean equals(Object o) {
      if (!(o instanceof Pair)) return false;
      Pair pairo = (Pair) o;
      return X.equals(pairo.getX()) && Y.equals(pairo.getY());
    }
  }
  
  static final long INFL = 200000000000000000L;
  static final int INF = 2000000000;
  static final boolean DEBUG = true;
  
  static StreamTokenizer in;
  static PrintWriter out;
  
  static void print(String s) {
    print(s, 0);
  }
  
  static void print(String s, int debug) {
    if (debug == 0 || DEBUG) {
      out.print(s);
    }
  }
  
  static void println(String s) {
    println(s, 0);
  }
  
  static void println(String s, int debug) {
    if (debug == 0 || DEBUG) {
      out.println(s);
    }
  }
  
  static void printArray(int[] arr) {
    println(Arrays.toString(arr));
  }
  
  static void printArray(char[] arr) {
    println(Arrays.toString(arr));
  }
  
  static void printArray(String[] arr) {
    println(Arrays.toString(arr));
  }
  
  static void sort(int[] a) {
    Random rnd = new Random();
    for (int i = a.length - 1; i > 0; i--) {
      int index = rnd.nextInt(i);
      a[i] ^= a[index];
      a[index] ^= a[i];
      a[i] ^= a[index];
    }
    Arrays.sort(a);
  }
  
  static char[] inChars;
  static int inCharsInd;
  
  static String next() throws IOException {
    in.nextToken();
    return in.sval;
  }
  
  static char nextChar() throws IOException {
    while (inChars == null || inCharsInd >= inChars.length) {
      inChars = next().toCharArray();
      inCharsInd = 0;
    }
    return inChars[inCharsInd++];
  }
  
  static int nextInt() throws IOException {
    return Integer.valueOf(next());
  }
  
  static double nextDouble() throws IOException {
    return Double.valueOf(next());
  }
  
  static long nextLong() throws IOException {
    return Long.valueOf(next());
  }
  
  private void run() throws IOException {
    solve();
  }
  
  int K, L, M, N, P;

  class Vertex{
    ArrayList<Path> neighbors = new ArrayList<>();
    boolean isStock = false;
  }

  class Path{
    Vertex v;
    long cost;
    public Path(Vertex v, int cost) {
      this.v = v;
      this.cost = cost;
    }
  }

  private void solve() throws IOException {
    N = nextInt();
    M = nextInt();
    K = nextInt();
    Vertex[] vertices = new Vertex[N];
    for (int i = 0; i < N; i++) {
      vertices[i] = new Vertex();
    }
    for (int i = 0; i < M; i++) {
      int a = nextInt() - 1;
      int b = nextInt() - 1;
      int cost = nextInt();
      vertices[a].neighbors.add(new Path(vertices[b], cost));
      vertices[b].neighbors.add(new Path(vertices[a], cost));
    }
    
    long minCost = INFL;
  
    for (int i = 0; i < K; i++) {
      vertices[nextInt()-1].isStock = true;
    }


    for (int i = 0; i < N; i++) {
      Vertex vertex = vertices[i];
      if (vertex.isStock){
        for (Path p: vertex.neighbors) {
          if (!p.v.isStock){
            minCost = Math.min(minCost, p.cost);
          }
        }
      }
    }
    
    if (minCost == INFL) minCost = -1;
    
    out.println(minCost);
  }
  
}
