#include<bits/stdc++.h>
using namespace std;
ifstream fin("team.in");
ofstream fout("team.out");
typedef pair<int,int> str;
const int DIM = 505;
const int INF = 0x3f3f3f3f;
int P, N, M, Dist[55][DIM], C[DIM][DIM], Dp[55][55][55], Home[55];
vector<int> Edge[DIM];
priority_queue< str, vector<str>, greater<str> > q;
inline void dijkstra( int pos ){
    memset( Dist[pos], INF, sizeof(Dist[pos]) );
    Dist[pos][ Home[pos] ] = 0;
    q.push( { 0, Home[pos] } );
    while( !q.empty() ){
        int nod = q.top().second;
        int Lg = q.top().first;
        q.pop();
        if( Dist[pos][nod] != Lg )
            continue;
        for( int i = 0; i < Edge[nod].size(); i++ ){
            int vecin = Edge[nod][i];
            if( Dist[pos][vecin] > Dist[pos][nod] + C[nod][vecin] )
                Dist[pos][vecin] = Dist[pos][nod] + C[nod][vecin], q.push( { Dist[pos][vecin], vecin } );
        }
    }
}
int solve( int i, int j, int k ){
    if( i > j )
        return 0;
    if( Dp[i][j][k] != INF )
        return Dp[i][j][k];
    for( int cross = i; cross <= j; cross++ ){
        //formez grupul i..j reunind grupurile i..cross-1 si cross+1...j cu persoana de legatura cross in orasul acesteia
        Dp[i][j][k] = min( Dp[i][j][k], Dist[k][ Home[cross] ] +
                                      solve( i, cross - 1, cross ) + solve( cross + 1, j, cross ) );
    }
    return Dp[i][j][k];
}
int main(){
    fin >> P >> N >> M;
    for( int i = 1; i <= M; i++ ){
        int a, b; fin >> a >> b;
        fin >> C[a][b]; C[b][a] = C[a][b];
        Edge[a].push_back( b );
        Edge[b].push_back( a );
    }
    for( int i = 1; i <= P; i++ )
        fin >> Home[i];
    Home[++P] = 1;
    sort( Home + 1, Home + P + 1 );
    int K = 1;
    for( int i = 2; i <= P; i++ )
        if( Home[K] != Home[i] )
            Home[++K] = Home[i];
    P = K;
    for( int i = 1; i <= P; i++ )
        dijkstra( i );
    //Dp[i][j][k] = costul minim pentru ca persoanele i...j sa ajunga acasa pornind din orasul persoanei k
    memset( Dp, INF, sizeof(Dp) );
    for( int i = 1; i <= P; i++ )
        Dp[i][i][i] = 0;
    fout << solve( 1, P, 1 ) << "\n";
    return 0;
}