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 com.sun.org.apache.xerces.internal.util.SynchronizedSymbolTable;

import java.awt.*;
import java.io.*;
import java.lang.reflect.Array;
import java.util.*;
import java.util.List;

public class Newbie {

    static InputReader sc = new InputReader(System.in);
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        solver s = new solver();
        int t = 1;
        while (t > 0) {
            s.solve();
            t--;
        }
    }

    static class descend implements Comparator<pair1> {
        public int compare(pair1 o1, pair1 o2) {
            if (o1.pop != o2.pop)
                return (int) (o1.pop - o2.pop);
            else
                return o1.in - o2.in;
        }
    }

    static class InputReader {
        public BufferedReader br;
        public StringTokenizer token;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream), 32768);
            token = null;
        }

        public String next() {
            while (token == null || !token.hasMoreTokens()) {
                try {
                    token = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return token.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }

    static class card {
        long a;
        int cnt;
        int i;

        public card(long a, int cnt, int i) {
            this.a = a;
            this.cnt = cnt;
            this.i = i;
        }
    }

   /* static class ascend implements Comparator<pair> {
        public int compare(pair o1, pair o2) {
            if (o1.a != o2.a)
                return (int) (o1.a - o2.a);
            else
                return (int) (o1.b - o2.b) * -1;
        }
    }*/

    static class extra {

        static boolean v[] = new boolean[100001];
        static List<Integer> l = new ArrayList<>();
        static int t;

        static void shuffle(long a[]) {
            List<Long> l = new ArrayList<>();
            for (int i = 0; i < a.length; i++)
                l.add(a[i]);
            Collections.shuffle(l);
            for (int i = 0; i < a.length; i++)
                a[i] = l.get(i);
        }

        static long gcd(long a, long b) {
            if (b == 0)
                return a;
            else
                return gcd(b, a % b);
        }

        static boolean valid(int i, int j, int r, int c) {
            if (i >= 0 && i < r && j >= 0 && j < c)
                return true;
            else
                return false;
        }

        static void seive() {
            for (int i = 2; i < 100001; i++) {
                if (!v[i]) {
                    t++;
                    l.add(i);
                    for (int j = 2 * i; j < 100001; j += i)
                        v[j] = true;
                }
            }
        }

        static int binary(long a[], long val, int n) {
            int mid = 0, l = 0, r = n - 1, ans = 0;
            while (l <= r) {
                mid = (l + r) >> 1;
                if (a[mid] == val) {
                    r = mid - 1;
                    ans = mid;
                } else if (a[mid] > val)
                    r = mid - 1;
                else {
                    l = mid + 1;
                    ans = l;
                }
            }
            return (ans + 1);
        }

        static long fastexpo(int x, int y) {
            long res = 1;
            while (y > 0) {
                if ((y & 1) == 1) {
                    res *= x;
                }
                y = y >> 1;
                x = x * x;
            }
            return res;
        }

        static long lfastexpo(int x, int y, int p) {
            long res = 1;
            x = x % p;
            while (y > 0) {
                if ((y & 1) == 1) {
                    res = (res * x) % p;
                }
                y = y >> 1;
                x = (x * x) % p;
            }
            return res;
        }
    }

    static class pair {
        int v;
        long w;

        public pair(int a, long n) {
            this.v = a;
            this.w = n;
        }
    }

    static class pair1 {
        long pop;
        int in;

        public pair1(long a, int n) {
            this.pop = a;
            this.in = n;
        }
    }

    static long m = (long) 1e9 + 7;

    static class solver {
        void solve() {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int k = sc.nextInt();
            List<pair> l[] = new ArrayList[n+1];
            for (int i = 0; i < n+1; i++)
                l[i] = new ArrayList<>();
            for (int i = 0; i < m; i++) {
                int u = sc.nextInt();
                int v = sc.nextInt();
                long w = sc.nextLong();
                l[u].add(new pair(v, w));
                l[v].add(new pair(u, w));
            }
            List<Integer> supp = new ArrayList<>();
            int sup[] = new int[100005];
            for (int i = 0; i < k; i++) {
                int t = sc.nextInt();
                supp.add(t);
                sup[t] = 1;
            }
            if (k == 0)
                System.out.println(-1);
            else {
                long min = Long.MAX_VALUE;
                for (int i = 0; i < k; i++) {
                    int s = supp.get(i);
                    for (int j = 0; j < l[s].size(); j++) {
                        long w = l[s].get(j).w;
                        int v = l[s].get(j).v;
                        if (sup[v] != 1) {
                            min = Math.min(w, min);
                        }
                    }
                }
                if (min == Long.MAX_VALUE)
                    System.out.println(-1);
                else
                    System.out.println(min);
            }
        }
    }
}