#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 100001;
int tab[N][1<<5], dist[5][N];
vector<pair<int, int>> adj[N];
signed main() {
cin.sync_with_stdio(0);
cin.tie(0);
int n, k, m, ans = LLONG_MAX;
cin >> n >> k >> m;
vector<int> term(k);
for (int i = 0; i < k; i++) cin >> term[i];
for (int i = 0; i < m; i++) {
int a, b, c;
cin >> a >> b >> c;
adj[a].push_back({b, c});
adj[b].push_back({a, c});
}
for (int i = 0; i < k; i++) {
fill(dist[i], dist[i] + N, LLONG_MAX);
dist[i][term[i]] = 0;
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<>> q;
q.push({0, term[i]});
while (!q.empty()) {
auto [c, v] = q.top();
q.pop();
if (c != dist[i][v]) continue;
for (auto& [nv, w] : adj[v]) {
if (dist[i][v] + w < dist[i][nv]) {
dist[i][nv] = dist[i][v] + w;
q.push({dist[i][nv], nv});
}
}
}
}
for (int v = 1; v <= n; v++) adj[v].push_back({v, 0});
for (int i = 0; i < k; i++) {
for (int j = 0; j < N; j++) fill(tab[j], tab[j] + (1 << 5), LLONG_MAX);
tab[term[i]][1 << i] = 0;
priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>, greater<>> q;
q.push({0, term[i], 1 << i});
while (!q.empty()) {
auto [c, v, mask] = q.top();
q.pop();
// cout << c << " " << v << " " << mask << "\n";
if (c != tab[v][mask]) continue;
if (mask == (1 << k) - 1) {ans = min(ans, c); break;}
for (auto& [nv, w] : adj[v]) {
for (int t = 0, amount = 0; t < k; t++) {
int nmask = mask | (1<<t);
if (nmask > mask) amount = dist[t][nv];
if (c + w + amount < tab[nv][nmask]) {
tab[nv][nmask] = c + w + amount;
q.push({tab[nv][nmask], nv, nmask});
}
}
}
}
}
cout << ans << "\n";
return 0;
}
/*
resource: https://www.youtube.com/watch?v=BG4vAoV5kWw&t=499s
for k <= 5, >0 caterpillar graph possible
can do disjtra w/ mask and node state
tle + bad alloc. instead of extra 2^k factor edges
ensure can only retreive one terminal at a time, but rego to self for multiple terminal lines
so its only extra k factor edges
tle last batck, k = 5. hmmm...
*/
