//#include <bits/stdc++.h>

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cstring>
#include <complex>
#include <stack>
#include <queue>
#include <unordered_map>
#include <map>
using namespace std;

struct UnionFind{
   vector<long long int> par; //par[i] : iの親の番号
   vector<long long int> rnk; //root[i] : iの木のランク

   //コンストラクタ
   UnionFind(long long int n): par(n), rnk(n){
       //木の初期化
       for(long long int i = 0; i < n; i++){
           par[i] = i;
           rnk[i] = 1;
       }
   }

   //データxが属する木の根を再帰で取得
   long long int root(long long int x){
       if(par[x] == x){
           return x;
       }else{
           return par[x] = root(par[x]);
       }
   }

   //xとyが同じ木に属しているかを判定
   bool same(long long int x, long long int y){
       return root(x) == root(y);
   }

   //xとyの木を併合
   void unite(long long int x, long long int y){
       long long int rx = root(x); //xの根
       long long int ry = root(y); //yの根
       if(rx == ry) return; //根が同じならそのまま

       if(rnk[rx] < rnk[ry]){
           par[rx] = ry;
       }else{
           par[ry] =rx;
           if(rnk[rx] == rnk[ry]) rnk[rx]++;
       }
   }
};

int main(){
    
    long long  i, j, n, m, k, from, to, cost, a1, a2, a3;

    cin >> n >> m >> k;
    UnionFind uf(n);

    vector<pair<long long, pair<long long , long long> > > tmp_p(m);
    vector<pair<long long, pair<long long , long long> > > p(0);
    long long used[m];
    long long sumcost = 0;
    long long ans = 0;
    for(i=0;i<m;i++) used[i] = 0;
    
    for(i=0;i<m;i++) {
        cin >> from >> to >> cost;
        tmp_p[i] = make_pair(cost, make_pair(from-1, to-1));
        sumcost += cost;
    }

    for(i=0;i<k;i++) {
        cin >> a1;
        //cout << a1 << endl;
        used[a1-1] = 1;
    }

    for(i=0;i<m;i++) {
        //scout << used[i] << endl;
        if(used[i]) {
            uf.unite(tmp_p[i].second.first, tmp_p[i].second.second);
            ans += tmp_p[i].first;
        } else {
            p.push_back(tmp_p[i]);
        }
    }

    sort(p.begin(), p.end());
    
    
    for(i=0;i<p.size();i++){
        //cost = first
        //from = second.fist
        //to   = second.second
        if(!uf.same(p[i].second.first, p[i].second.second)){
            uf.unite(p[i].second.first, p[i].second.second);
            ans += p[i].first;
        }
    }
    /*cout << sumcost << endl;
    cout << ans << endl;*/
    cout << sumcost-ans << endl;

}