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

using namespace std;
using namespace atcoder;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
static const double pi = 3.141592653589793;
const ll INF = 1LL << 60;
const ll mod = 1000000007;
const ll imod = 998244353;
using mint = modint998244353;
vector<int> dx = {1, 0, -1, 0}, dy = {0, 1, 0, -1};
ll P(ll x, ll n) {
  ll ret = 1;
    while (n > 0) {
      if (n & 1) ret *= x;
      x *= x;
      n >>= 1;
  }
  return ret;
}

void seek(bool f){
  cout << (f ? "Yes" : "No") << endl;
}

int main(){
	int N, M, K;
	cin >> N >> M >> K;
	vector<int> OK(N, 0);
	rep(i, K){
		int x;
		cin >> x;
		x--;
		OK[x] = 1;
	}
	vector<vector<int>> T(N, vector<int> (N));
	rep(i, N){
		rep(j, N){
			cin >> T[i][j];
		}
	}
	vector<int> order(N, 0);
	rep(i, M){
		order[i] = 1;
	}
	sort(order.begin(), order.end());
	int ans = 1e9;
	do{
		vector<int> A;
		rep(i, N){
			if(order[i] == 1){
				A.push_back(i);
			}
		}
		do{
			int res = 0;
			rep(j, M - 1){
				res += T[A[j]][A[j + 1]];
			}
			if(OK[A[M - 1]] == 0){
				int add = 1e9;
				rep(k, N){
					if(OK[k] == 1){
						add = min(add, T[A[M - 1]][k]);
					}
				}
				res += add;
			}
			ans = min(ans, res);
		}while(next_permutation(A.begin(), A.end()));
	}while(next_permutation(order.begin(), order.end()));
	cout << ans << endl;
}