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.StringTokenizer;

public class B {

    public static void main(String[] args) throws IOException {


        new B().solve();
    }

    int mod = 1000000007;

    PrintWriter out;

    int n;
    int m;

    int inf = 2000000000;

    void solve() throws IOException {

        Reader in;

        try {

            in = new Reader("input.txt");
            out = new PrintWriter( new BufferedWriter(new FileWriter("output.txt")) );
        }
        catch (Exception e) {

            in = new Reader();
            out = new PrintWriter( new BufferedWriter(new OutputStreamWriter(System.out)) );
        }

        //BufferedReader br = new BufferedReader( new FileReader("input.txt") );
        //BufferedReader br = new BufferedReader( new InputStreamReader( System.in ) );

        int n = in.nextInt();
        int m = in.nextInt();
        int k = in.nextInt();

        ArrayList<Pair>[] g = new ArrayList[n];

        for (int i = 0; i < n; i++)
            g[i] = new ArrayList<Pair>();

        for (int i = 0; i < m; i++) {

            int x = in.nextInt()-1;
            int y = in.nextInt()-1;
            int l = in.nextInt();
            g[x].add(new Pair(y, l));
            g[y].add(new Pair(x, l));
        }

        int[] b = new int[k];
        int[] pek = new int[n+1];
        for (int i = 0; i < k; i++) {

            int x = in.nextInt()-1;
            b[i] = x;
            pek[x] = 1;
        }

        int ans = inf;

        for (int i = 0; i < k; i++) {

            int x = b[i];
            for (Pair u : g[x])
                if (pek[u.a] == 0)
                    ans = Math.min(ans, u.b);
        }

        if (ans == inf)
            ans = -1;

        out.println(ans);
        out.flush();
        out.close();
    }


    //class Edge { int v, u; int w, c; Edge(int v, int u, int w, int c) { this.v = v; this.u = u; this.w = w; this.c = c; } }

    class Pair implements Comparable<Pair>{

        int a;
        int b;

        Pair(int a, int b) {

            this.a = a;
            this.b = b;
        }

        public int compareTo(Pair p) {


            return 0;
        }
    }

    class Item {

        int a;
        int b;
        int c;

        Item(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }

    }

    class Reader {

        BufferedReader  br;
        StringTokenizer tok;

        Reader(String file) throws IOException {
            br = new BufferedReader( new FileReader(file) );
        }

        Reader() throws IOException {
            br = new BufferedReader( new InputStreamReader(System.in) );
        }

        String next() throws IOException {

            while (tok == null || !tok.hasMoreElements())
                tok = new StringTokenizer(br.readLine());
            return tok.nextToken();
        }

        int nextInt() throws NumberFormatException, IOException {
            return Integer.valueOf(next());
        }

        long nextLong() throws NumberFormatException, IOException {
            return Long.valueOf(next());
        }

        double nextDouble() throws NumberFormatException, IOException {
            return Double.valueOf(next());
        }

        String nextLine() throws IOException {
            return br.readLine();
        }

    }

    class Parser {

        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public Parser(InputStream in) {
            din = new DataInputStream(in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead =  0;
        }
        public String nextString(int maxSize) {
            byte[] ch = new byte[maxSize];
            int point =  0;
            try {
                byte c = read();
                while (c == ' ' || c == '\n' || c=='\r')
                    c = read();
                while (c != ' ' && c != '\n' && c!='\r') {
                    ch[point++] = c;
                    c = read();
                }
            } catch (Exception e) {}
            return new String(ch, 0,point);
        }
        public int nextInt() {
            int ret =  0;
            boolean neg;
            try {
                byte c = read();
                while (c <= ' ')
                    c = read();
                neg = c == '-';
                if (neg)
                    c = read();
                do {
                    ret = ret * 10 + c - '0';
                    c = read();
                } while (c > ' ');

                if (neg) return -ret;
            } catch (Exception e) {}
            return ret;
        }
        public long nextLong() {
            long ret =  0;
            boolean neg;
            try {
                byte c = read();
                while (c <= ' ')
                    c = read();
                neg = c == '-';
                if (neg)
                    c = read();
                do {
                    ret = ret * 10 + c - '0';
                    c = read();
                } while (c > ' ');

                if (neg) return -ret;
            } catch (Exception e) {}
            return ret;
        }
        private void fillBuffer() {
            try {
                bytesRead = din.read(buffer, bufferPointer =  0, BUFFER_SIZE);
            } catch (Exception e) {}
            if (bytesRead == -1) buffer[ 0] = -1;
        }

        private byte read() {
            if (bufferPointer == bytesRead) fillBuffer();
            return buffer[bufferPointer++];
        }
    }

}