/*
1. 状态定义： 花费最小 -> 什么状态花费最小？ -> 服务员在x,y,z 时 f[x][y][z]最小 
            -> 请求来源是线性的 -> 处理完第i请求时的花费是可以递推的 -> 处理完1..i 个请求时 f[i][x][y][z] 最小
            -> 处理第i个时必定有一个员工在i处 -> 即求处理完1..i 个请求时的代价f[i][a][b]
2. 状态计算： 

2.1 状态转移：“把每个状态看成一张表，DP就是求最短路”，根据每次指派到第i+1个点的情况向后递推
            f[i][a][b] = MIN(f[i][a][b] , f[i-1][a][b] + cost(p[i-1], p[i]))

2.2 合法性计算：     if (x == y || x == z || y == z) continue;

2.3 边界条件： 假定０号任务由任意ｐ[0] 去，故设f[0][2][3] = 0, p[0] = 1 
                取MIN所以初始化为INF
                需要设定0， 所以哨兵也要初始化为INF

*/


import java.util.*;

public class Main{

    void run(){
        int l = jin.nextInt();
        int n = jin.nextInt();
        for (int i = 1 ; i <= l; i++) 
            for (int j = 1; j <= l; j++)
                cost[i][j] = jin.nextInt();
        for (int i = 1 ; i <= n ; i++) p[i] = jin.nextInt();

        for (int i = 0 ; i <= n ; i++){  // 初始化从0 开始
            for (int x = 0; x <= l ; x++){
                for (int y = 0 ; y <= l ; y++){
                    f[i][x][y] = Integer.MAX_VALUE >> 1;
                }
            }
        }
        f[0][2][3] = 0;
        p[0] = 1;

        // for (int i = 1 ; i <= n ; i++){
        for (int i = 0 ; i < n ; i++){
            for (int x = 1; x <= l ; x++){
                for (int y = 1 ; y <= l ; y++){
                    int z = p[i];
                    if (x == y || x == z || y == z) continue;

                    int v = f[i][x][y];
                    int next = p[i+1];
                    f[i+1][x][y] = Math.min(f[i+1][x][y], v + cost[z][next]);
                    f[i+1][x][z] = Math.min(f[i+1][x][z], v + cost[y][next]);
                    f[i+1][y][z] = Math.min(f[i+1][y][z], v + cost[x][next]);
                }
            }
        }

        int res = Integer.MAX_VALUE;
        for (int x = 1 ; x <= l ; x++){
            for (int y = 1; y <= l; y++){
                int z = p[n];
                if (x == y || x == z || y == z) continue;
                res = Math.min(res, f[n][x][y]);
            }
        }
        System.out.println(res);

    }

    int maxN = 1002;
    int maxL = 202;
    int[] p = new int[maxN];
    int[][][] f = new int[maxN][maxL][maxL];
    int[][] cost = new int[maxL][maxL];
    Scanner jin = new Scanner(System.in);
    public static void main(String[] args) {new Main().run();}
}