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. 
//package codes;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.stream.Collectors;

//import codes.Reference.i;

public class Bakery {
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		long t[];
		
		t = i.la();
		long n = t[0];
		long m = t[1];
		long k = t[2];
		//String s = i.s();
		ArrayList<uvl> arr = new ArrayList<uvl>();
		while(m>0)
		{
			long a[]=i.la();
			arr.add(new uvl(a[0],a[1],a[2]));
			m--;	
		}
		
		long storage[]= {};
		if(k!=0)
		{
			 storage = i.la();
		}
		
		Collections.sort(arr,new Comparator<uvl>() {

			@Override
			public int compare(uvl o1, uvl o2) {
				// TODO Auto-generated method stub
				return Long.compare(o1.l, o2.l);
			}
			
		});
		
		HashMap<Long, Integer> map = new HashMap<Long, Integer>();
		
		for(int i=0;i<storage.length;i++)
		{
			map.put(storage[i], 0);
		}
		int f=0;
		uvl uv = null;
		for(uvl o:arr)
		{
			if((map.containsKey(o.u) && !map.containsKey(o.v)) || (!map.containsKey(o.u) && map.containsKey(o.v)))
			{
				f=1;
				uv=o;
				break;
			}
		}
		if(f==1)
		{
			System.out.println(uv.l);
		}
		else
			System.out.println(-1);
		
	}
	static private class i
	{
		//input taking methods
		static BufferedReader b = new BufferedReader(new InputStreamReader(System.in));
		
		public static String[] sa()
	    {
	    	try {
				return b.readLine().split(" ");
			} catch (IOException e) {
				
				e.printStackTrace();
			}
	    	return null;
	    }
		
		public static long l()
		{
			try {
				return Long.parseLong(b.readLine());
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
	    	return 0;
		}
		
		public static String s()
		{
			try {
				return b.readLine();
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
	    	return null;
		}
		
		public static long[] la()
		{
			return Arrays.stream(sa()).parallel().mapToLong(Long::parseLong).toArray();
		}
		
		public static ArrayList<Long> lal()
		{
			return (ArrayList<Long>)Arrays.stream(la()).parallel().boxed().collect(Collectors.toList());
		}
		public static StringBuilder so(String a)
		{
			return new StringBuilder(a);
		}

	}
	/*
	 HashMap<String,Integer> map = new HashMap<String,Integer>();
	 if(!map.containsKey(key))
	 {
	 	map.put(key,value);
	 }
	 else
	 {
	 	map.replace(key,map.get(key));
	 } 
	 */
	static private class algo{
		
		public static boolean isPrime(int a)
		{
			for(int i=2;i<a;i++)
			{
				if(a%i==0)
				{
					return false;
				}
			}
			return true;
		}
		static String reverse(String s)
		{
			return i.so(s).reverse().toString();
		}
		static String replace(String s,int st,int e,String repl)
		{
			return i.so(s).replace(st, e, repl).toString();
		}
		static String insert(String s,int pos,String ins)
		{
			return i.so(s).insert(pos, ins).toString();
		}
		static long sod(long n) 
	    { 
	        long sum; 
	        for (sum = 0; n > 0; sum += n % 10, 
	                                  n /= 10); 
	  
	        return sum; 
	    } 

		static int gcd(int a, int b) 
	    { 
			if (a == 0) 
				return b;  
			return gcd(b % a, a);  
	    } 
		
	    static int lcm(int a, int b) 
	    { 
	        return (a*b)/gcd(a, b); 
	    }
	    static long sumOfSquares(long[] a)
	    {
	    	
	    	return (long) Arrays.stream(a).mapToDouble(i->i*i).sum();
	    }
	    //use of filter
	    //Arrays.stream(a).filter(k->k==1).count();
	    //            array       variable->condition returning boolean  . lot fof functionsa re available
	    
	    static String[] generateBinSeq(int cnt)
	    {
	    	int loopvar=(int) Math.pow(2, cnt);
			String a[][] = new String[cnt+1][loopvar];
			
			for(int i=1;i<=cnt;i++)
			{
				int temp = (int) Math.pow(2, i);
				for(int j=0;j<temp;j++)
				{
					if(i==1 || i==2)
					{
						if(i==1)
						{
							a[i][0]="0";
							a[i][1]="1";
							continue;
						}
						else
						{
							a[2][0]="00";a[2][1]="01";a[2][2]="10";a[2][3]="11";
							continue;
						}
					}
					else
					{
						if(j<(temp/2))
						{
							a[i][j] = a[i-1][j]+"0"; 
						}
						else
						{
							a[i][j] = a[i-1][j%((temp+1)/2)]+"1";
						}
				
					}
						
				}
			}
	    
			return a[cnt];
	    }
	    
	 // USe this snippet for calling
//	    int ndx = algo.bs(a,key,0,a.length-1);
//		if(ndx<0)
//			ndx=0;
//		if(ndx==a.length)
//			ndx=a.length-1;

	    static public int bs(long a[],long key,int s,int e)
		{
			int m = (s+e)/2;
			if(s>e)
				return s;
				
			if(a[m]==key)
				return m;
			else if(a[m]>key)
				return bs(a,key,s,m-1);
			else
				return bs(a,key,m+1,e);
			
		}
	 // USe this snippet for calling
//	    int ndx = bsa(al,key,0,al.size()-1);
//		if(ndx<0)
//			ndx=0;
//		if(ndx==al.size())
//			ndx=al.size()-1;
	    static public int bsa(ArrayList<Long> a,long key,int s,int e)
		{
			int m = (s+e)/2;
			if(s>e)
				return s;
				
			if(a.get(m)==key)
				return m;
			else if(a.get(m)>key)
				return bsa(a,key,s,m-1);
			else
				return bsa(a,key,m+1,e);
			
		}
	    static public long fact(long a)
	    {
	    	long i,fact=1;  
	    	 long number=a;   
	    	  for(i=1;i<=number;i++){    
	    	      fact=fact*i;    
	    	  } 
	    	  return fact;
	    }
	    
	}
}
class uvl{
	long u;
	long v;
	long l;
	public uvl(long u, long v, long l) {
		this.u = u;
		this.v = v;
		this.l = l;
	}
}
