#include <cstdio>
#include <vector>
#include <map>
#include <set>
#include <cmath>
#include <algorithm>

#define NMax 2001
#define Inf 1<<30

using namespace std;

long n, m, k, x, y, z, now, nr, minNr, maxK, sq;
long a[21], cost[21][NMax], pw2[21];
vector<pair<long, long> > edge[NMax];

void computeMinimalRoadFrom ( long ind, long now ) {
    set<pair<long, long> > q;

    for ( long i = 1; i <= n; i++ )
        cost[ind][i] = Inf;

    q.insert ( make_pair ( 0, now ) );
    cost[ind][now] = 0;
    while ( !q.empty ( ) ) {
        now = q.begin ( )->second;
        q.erase ( q.begin ( ) );

        for ( long i = 0; i < edge[now].size ( ); i++ )
            if ( cost[ind][now] + edge[now][i].second < cost[ind][edge[now][i].first] )
                cost[ind][edge[now][i].first] = cost[ind][now] + edge[now][i].second,
                q.insert ( make_pair ( cost[ind][edge[now][i].first], edge[now][i].first ) );
    }
}

long optimal ( long last, long sequence );

int main()
{
    freopen ( "ubuntzei.in", "r", stdin );
    freopen ( "ubuntzei.out", "w", stdout );

    scanf ( "%ld %ld %ld", &n, &m, &k );
    maxK = (1 << k) - 1; pw2[0] = 1;
    for ( long i = 1; i <= k; i++ ) {
        pw2[i] = pw2[i - 1] * 2;
        scanf ( "%ld", &a[i] );
    }

    for ( long i = 1; i <= m; i++ )
        scanf ( "%ld %ld %ld", &x, &y, &z ),
        edge[x].push_back ( make_pair ( y, z ) ),
        edge[y].push_back ( make_pair ( x, z ) );


    a[0] = 1;a[k + 1] = n;
    for ( long i = 0; i <= k + 1; i++ )
        computeMinimalRoadFrom ( i, a[i] );

    minNr = Inf;
    if ( k == 0 )
        minNr = cost[0][n];
    else
        for ( long i = 1; i <= k; i++ )
            minNr = min ( minNr, optimal ( i, maxK ) + cost[i][n] );

    printf ( "%ld\n", minNr );

    return 0;
}


long optimal (  long last, long sequence ) {
    long lg2 = log ( sequence ) / log ( 2 );
    if ( pw2[lg2] == sequence )
        return cost[0][a[k - lg2]];

    sequence -= pw2[k - last];
    long minNrTmp = Inf;
    for ( long i = 0, st = sequence, pw = 1; i < k; i++, st = st >> 1, pw = pw << 1 )
        if ( st & 1 )
            minNrTmp = min ( minNrTmp,  optimal( k - i, sequence ) + cost[k - i][a[last]] );

    return minNrTmp;
}